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# TOP IN CONTEMP MATH II MATH 166

Texas A&M

GPA 3.92

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This 114 page Class Notes was uploaded by Evert Christiansen on Wednesday October 21, 2015. The Class Notes belongs to MATH 166 at Texas A&M University taught by D. Lackey in Fall. Since its upload, it has received 12 views. For similar materials see /class/226013/math-166-texas-a-m-university in Mathematics (M) at Texas A&M University.

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Math 166 Lecture Notes 23 Section 23 Use of Counting Techniques in Probability Computing the Probability of an Event in a Uniform Sample Space Let S be a uniform sample space and let E be any event Then Number of favorable outcomes in E nE Number of possible outcomes in S nS PE When we rst began using this formula the numbers in E and the numbers in S were small enough to count easily As the sample spaces increase in size the problems increase in dif culty because the counting gets more dif cult Now that we have learned several counting techniques permutations combinations and the multiplication principle we can use these techniques to help us count nE and nS in more dif cult problems Give answers rounded to four decimal places Example 1 Two cards are selected at random from a wellshuffled pack of 52 playing cards What is the probability that a They are both diamonds HE nE Number of ways to get 2diamonds nS Number of ways to get any 2 cards b Neither of them is a diamond HE nE Number of ways to get 2 non diamonds nS Number of ways to get any 2 cards Example 2 Andy needs four studs for a small dog house At the lumber yard there is a stack oftwenty studs seven of which are crooked If Andy randomly grabs five studs from the stack to purchase and take home what is the probability that he will have enough straight studs for his project nE Number of ways to get at least 4 straight studs P E 7 nS Number of ways to get any 5 studs of ways to get 4 straight studs and lcrooked one of ways to get 5 straight studs Number of ways to get any 5 studs Example 3 Ben after studying for a spelling test knows how to spell 15 ofthe 20 words on the list lfthe teacher selects 10 words forthe quiz what is the probability that Ben Will make an A 90 or above on the quiz assume all words count equal points HE nE Number of ways to get at least words correct nS Number of ways to pick any 10 words for the quiz of ways to get exactly 9 correct amp 1 wrong of ways to get all 10 correct Number of ways to pick any 10 words for the quiz Example 4 An unbiased 3color spinner red blue green is spun four times What is the probability that the spinner will land a On red exactly three times nE Number of ways to land on red 3 of the 4 times and something else the 4th time P E nS Number of ways to spin 4 times and land on anything b On red at most two times HE nE Number of ways to land on red at most two times nS Number of ways to spin 4 times and land on anything of ways to land on red 0 times of ways to land on red once ways to land on red twice Number of ways for the spinner to land in four spins c On red on the first and the third spin nE Number of ways to land on red the 1st and 3rd spin P E nS Number of ways to spin 4times and land on anything d On red twice or green twice nE ways to get red twice ways to get green twice ways to get green twice amp red twice P E nS Number of ways to spin 4 times and land on anything Example 5 Mice are shipped out in boxes of 18 to a distributor Six mice out of every box are tested for a certain virus If any of the tested mice have the virus the entire box is quarantined What is the probability that a box of mice is shipped that actually has 2 mice infected with the virus nE Number of ways pull a 6mice sample of healthy mice P E 7 nS Number of ways to pull any 6 mice for a sample Note Since two mice are infected 16 are healthy Example 6 A modified Birthday Problem An office is celebrating October birthdays for ve people What is the probability that at least two of them have the same birthday Suggested approach Use rule of complements PE1 PEc The complement of at least two have the same birthday is no birthdays the samequot So we need to compute the probability that the birthdays are all different nEC 1 Number of ways to get 5 different birthdays nS Number of ways to get 5 quotanyquot birthdays 1lt gt1 1 1 1 gt1 PE1 PE 1 Example 6 EXTENSION For the general case involving yearround birthdays for r people if E denotes the event that at least two of the rpeople have the same birthday we have 365364363365 r1 365r We can choose some values for r and compute the associated probabilities Observe the results as shown in the table below PE 1 r5101520253035404550 PE 027 117 253 411 476 507 569 706 891 970 Notice that in a group of 50 people it is almost certain 97 probability that at least 2 people in the group will have the same birthday Whatis the probability that your Math section will have atleast two people with the same birthday Math 166 Lecture Notes F3 Section F3 Annuities and Sinking Funds An annuity is a sequence of payments made at regular time intervals The time period in which these payments are made is called the term of the annuity An annuity in which the payments are made at the end of each payment period is called an ordinary annuity An annuity in which the payments are made at the beginning of each period is called an annuity due An annuity in which the payment period coincides with the interest compounding period is called a simple annuity An annuity in which the payment period differs from the interest compounding period is called a complex annuity Note We will be considering annuities thathave terms with xed time intervals periodic payments equal in size payments made at the end of the paymentpen39ods and payment periods that coincide with the interest compounding period The future value 8 of an annuity of n payments of R dollars each paid at the end of each investment period into an account that earns interest at the rate ofi per period is SRW l The future value ofan annuity is easy to understand It is the value of the investment at the end of the investment period all the payments plus all the accumulated interest The present value P of an annuity of n payments of R dollars each paid at the end of each investment period into an account that earns interest at the rate ofi per period is l The present value ofan annuity might not be as clear in meaning We will explore this in one of our examples A sinking fund is any account that is established for accumulating funds to meet future obligations or debts There are different types of sinking funds One type of sinking fund would be to make payments in the form ofan ordinary annuity Forthis type we can apply the principles in this section of material We will be using the TVM Solver again rather than nance tables to solve for future values and present values of annuities and payment amounts We could also solve for interest rates and time periods Example 1 In order to save for David s college education his parents decide to deposit 200 at the end of every month into a bank account paying interest at the rate of 6 per year compounded monthly lfthe savings program begins when David is 2 years old how much money will have accumulated by the time he turns 17 N l PV PMT FV PY CY How much was investment and how much was interest Example 2 Find the future value of an ordinary annuity of 36 monthly payments of 150 earning interest at 85 per year compounded monthly N FV PV P Y pV PMT CY Now find the present value N FV FV PV P Y PMT CY What is the meaning of present value in a case such as this Use your computed present value rather than payments and compute the future value What do you observe N I FV PV PY FV PMT CY Example 3 Mr Johnson has a goal to have 400000 in a retirement account He plans to achieve that goal by investing a certain amount every month for the next 25 years into an ordinary annuity earning 675 compounded monthly How much does Mr Johnson s monthly payment need to be N I FV PV PY PMT CY Example 4 Cynthia is planning to make a contribution of 4000 on January 31 of each year into an IRA earning interest at an effective rate of 49 per year After she makes her 15th payment how much will she have in the IRA N I FV PV PY PMT CY Example 5 James and Peter are the same age From age 35 to 50 Peter deposits 300 at the end of each month into a taxfree retirement account Then he makes no withdrawals or further contributions James does not begin making deposits into a retirement account until age 45 He then makes deposits of 450 at the end of each month into the same type of account until age 60 Both accounts earn interest at the rate of 58 compounded monthly At age 60 who ends up with the bigger nest egg Peter Stage1 N FV PV PMT PY CY Stage2 N FV PV PMT PY CY Question How much of Peter s own money did he invest James N FV pV PMT PY CY Question How much of James s own money did he invest Note They both invested monthly for 15 years but Peter s contributions were smaller Still Peter ended up with the greater nest egg Lesson to be learned Invest early and leave it there Example 6 Sarah decides to purchase a car She makes a down payment of 3500 and finances the rest Her monthly payment is 250 per month for 48 months Her interest rate is 9 per year compounded monthly on the unpaid balance What was the selling price ofthe car How much will Sarah end up paying in interest charges N FV PV PY PMT CY Selling price of the car Amount actually paid for the car including interest Interest paid Math 166 Exam 2 A Reminder of some ofthe Concepts to Know Caution This is notintended to be used as an allinclusive study guide It is designed to be a general reminder of many of the various things we have studied this period Know how to use a variety of counting techniques to find a the number of ways something can occur b the probability of an event occurring These involve combinations permutations and the multiplication principle Know how to determine when an experiment can be classi ed as a binomial experiment Know how to use the binompdfand binomcdf functions to solve problems involving binomial probabilities Review PXgt6 PX 2 6 PXlt 5 P3lt X 11 Know the categories of random variables Know how to find all possible outcomes for a random variable as defined Know how to create a probability distribution these can involve counting principles tree diagrams etc Understand histograms and the relationship to probability distributions Know how to find EX Understand what it is Know how to find mean median mode standard deviation variance From a single list From a frequency table or probability distribution Know the basic characteristics of a continuous random variable distribution especially the normal distribution Know how to use normalcdf and invnorm to solve problems involving normal distributions Review PXgt6 and compare to the binomial PXgt6 Specifically be able to recognize invnorm situations Know how to use Chebychev s Inequality in working with functions that are not as nice as normal distributions Math 166 Lecture Notes 34 Section 34 The Normal Distribution Probabili Densi Functions p i and therefore Werefinite orobabiiiti distributions We couid represent tne probabiiity distribution in atabietorrn or Witn a nistograrn in tnis Sections We iook at probabiiity distributions associated Witn continuous randorn variabies We retertotnern as continuous orobabiiiti distributions Since tne outcomes are infinite representation in a tabie or Witn a nistograrn is not possibie instead tnere is a probabiiity density tunction xi associated Witn tne data a continuous randorn variabie nas tne foHoWing properties i fxi5nonnegativeforaiivaiuesofx 2 Tne area ottne region between tne grapn of fandmexraxis is equai to i am Normal Distributions Normai distributions are a speciai ciass of continuous probabiiity distributions Tne grapn of a norrnai distribution is beii snaped and is caiied a normal curve is u i iiiiined by i in a rneasure of centrai tendency and tne standard deviation is a rneasure of its spread aboutt e mean 9973quot ias A5quot 53273 i i i 1 urlcr urZEr War it m r 2 imp A horrhat curve has the touowthg charactertsttcs t The curve has a peak atx d 2 The curve ts Symmemc th respect to the verttcat hhe x d 3 thdehhttety m ether dtrecttoh he area UH ert e curve ts 68 27 or the area under the curve hes thhth 1 standard devtattoh or the mean that 5 betw err or oarrd d o 95 45 or the area hes Wthth 2 standard devtattohs or the mean and 99 73 or the area hes Wthth 3 standard devtattohs or the mean Nb t a 5 mean 2 mean 4 Computation u However the vames are hot eashy observed ash a hhtte dtstrtouttoh taoTe or htstograrh e heed to resort to taoTes or caTcuTatormhcttohs regtohs u to the en or a gwen vame Makes pracuce to Tearh to usethetaoTes Toper y The taoTes m a textbook appenth are based on the standard horrhaT probabmty dtsthouttoh The standard horrhaT curve tsthe horrhaT curve wtth a mean or zero to 0 otohetot ot ourvetsh t the Standard norma Normal cdffunct n n ms dassy We M be uswng a prooaonny funcnon that 5 bunt nto your ca cu ator Tne nonna cdffuncnon n our ca cu alor audns us to obtawn probabmnesfor any nonna e Tnefonnalfortnwsfuncnon 5 normalgdf e endpoint right endpoint p o Example 1 Let X be a n0nna random Vanab e thn rnean of 100 and Standard dev auon of 20 USan ne nonna cdffuncuon n your Ca cu ator nd ne Va ues be oW a P00 lt xlt M5 o Pgtlt lt MO c PX gt 80 Examplez Lethetne standard nonna vanab e Uswngtne nonna cdffuncnon n your caxcuxaton nnd the venues be ow a P434ltzlt231 o PZlt214 Examples A Who esa e our company seus 20 pound bags of our The bags actuaHy vary n Wewgnt Howeyen tne Wewgnta of the bags are nonnaHy dwstnbuted wnn a mean Wewgnt of 20 pounds and a standard devwanon of 5 ounces n a bag of our 5 se ected at random from the warenouse What 5 the probabmty that n Wewgns a more than 20 pounds 8 ounces o between 19 5 pounds and 20 5 pounds Solving in reverse wed knowme probabmty and Want to fwd the range of random vanade vames m mm 7 Theformatfortmsmncuon 5 invnorm PZltz p a Note H d 7 dwlaw d we tame back to me approp ate rows and co umns Examgle 4 Let 2 be the standard norma vanab e Fwd me va ue of z w z saus es a PZltz 9115 Example 5 Let X be a norma random Vamab e WM mean 12 and Std dev 32 Fwd he Va ue of Z W Z Saus es a PZ gt z 4876 Example 6 Let 2 be the standard norma vanab e Fwd me va ue of z w z saus es a PrzltZltz 6778 Example 7 Tne annuar rncornes of a group of senorarsnrp apphcants are normaHy drstrrbured wrtn a rnean of 41000 and a standard dewanon of 3800 wer the duesuons berow roundmg to the nearest nundred doHars a Tnere were 50 apphcants rorrnrs senorarsnrp r omy 2o apphcants can be awarded a Schwarsmp and they wm be awarded to those wrtn the owest rncornes then What rstne rnawrnurn mcomemat an apphcant courd have and quahfyfor a Schwarsmp h an of H1 800f Candrdate hst What 5 he CULOquot ncome that WM determme Whether an apphcant recewes a etter or not Math 166 Lecture Notes 51 Section 51 Matrices Using Matrices to Represent Data A matrix is an ordered rectangular array of numbers A matrix with m rows and n columns has size m X n The entry in the ith row and jth column is denoted by all or aw Example 1 The Excellent Eggs Family Farm markets their eggs in three sizes They have three locations where they house their hens and collect the eggs The farm s egg production for March is summarized in the table below Table 1 Excellent Eggs Family Farm March production Size S Size M Size L Uncle Joe39s 120 380 460 Aunt Sue39s 80 340 410 Cousin Bob39s 140 420 500 Use a matrix to represent the data What is the size of the matrix Find 51113 and give an interpretation of the number Find all and give an interpretation ofthe number Find the sum of the entries that make up row 1 and interpret the results Find the sum of the entries that make up column 3 and interpret the result Eguality of Matrices Two matrices are equal if they are the same size AND their corresponding entries are equaL Example 2 Tell whether the following matrices are equal 1 3 5 1 2 a and 3 4 5 3 243 41 2 41 2 73 b and731 50 504 00 Addition and Subtraction of Matrices lfA and B are two matrices of the same size 1 The sum A B is the matrix obtained by adding the corresponding entries in the two matrices M The difference A B is the matrix obtained by subtracting the corresponding entries in B from A entries in A minus the corresponding entries in B lfA and B are two matrices of different sizes you cannot compute the sum or the difference Laws for Matrix Addition fA B and C are matrices ofthe same size then the commutative and associative properties hold true 1 Commutative property A B B A i 36 3 SH 5 2 Associative property A B C A B C C M W1 3H 13M 25H 3 Example 3 The total output of Excellent Eggs for April is shown in Table 2 Table 2 Excellent Eggs Family Farm April production Size S Size M Size L Uncle Joe39s 140 360 480 Aunt Sue39s 90 380 380 Cousin Bob39s 100 370 450 Find the total output of the company for March and April Use matrices Can you interpret the various elements in the total output matrix output for March output for April total for 2 months Transposed Matrices lfA is an m x n matrix with elements a then the transpose ofA is the n x m matrix referred to as AT and it has elements a Example4 4 2 LetA 1 0 FindAT 6 7 Scalar Multiplication lfA is a matrix and c is a real number then the scalar product cA is the matrix obtained by multiplying each entry ofA by 0 Example 5 LetA 7 0 2 5 3 8 Find 3A Matrix Equations 4 0 3 4 1 3 4 7 Given A and B and C 1 5 6 0 2 8 2 10 7 3 Example 6 Find the matrix Xsatisfying the matrix equation X 20 3BT Example 7 Find the matrix Xsatisfying the matrix equation 2X B 3A Note We might want to first solve for X in the matrix equation 2X B 3A as if we did not have any specific matrices A and B in mind Math 166 Lecture Notes M1 Section M1 Markov Chains Exgtonng a Qrob em usan tree dwagrams Exam Ataunw u 60 ofthe Hunmw mu m r months ofthe uh m at 0 of A AA a m fthetrend Contwnues WhatWH the usage took We n 12 months D m A M m fthetrend Contwnues WhatWH the usage took We n 18 months D m A M m f the trend Contwnues What WM the usage took We n 24 months D m A M m A Markov Chain is a process in which the probability associated with the outcomes at any stage of the process depend only on the outcomes ofthe preceding stage The outcome at any stage in a Markov process is called the state of the experiment The outcome at the current stage is called the current state Transition Probabilities are the probabilities associated with the transition from one state to the next in a Markov process We give them a matrix representation a a a 11 T all azy2 azy3 where a represents the conditional probability Pstate i state 1 12 13 a a a 31 32 33 The initialstate distributions and all subsequent currentstate distributions are also given a matrix or vector representation The transition matrix and the current state matrix are used to find the NEXT state Doing this repeatedly creates a chain process X1TX0 X2TX1 X3TX2 X4TX3 etc By applying a little algebra and substitution we can see that X4 T X3 2T T XZ 2TTTX1 2T T T T XO 2T4 X0 and in general we should be able to see that Xm TmX0 Exploring the same problem using matrices and the Markov process Example 1 again At a university it is estimated that 60 of the commuter students drive to school and the other 40 take the bus The university is upgrading the bus system and expects bus usage to increase in the next 6 months It is projected that 20 of the current car drivers will switch to using the bus and 90 of the current bus users will continue to use the bus What will the bus and car usage look like in 6 months lfthe trend continues what will the usage look like in 12 months lfthe trend continues what will the usage look like in 18 months lfthe trend continues what will the usage look like in 24 months lfthe trend continues what will the usage look like in 6 years What will happen ifthe trend continues inde nitely Example 2 A town hasthree pizza places A B and C Ifa customer ordered from A then the next time heshe orders a pizza there is a 50 chance heshe will order from A again 30 from B and 20 from C Ifa customer ordered from B then the next time heshe ordered a pizza there is a 40 chance heshe will order from A 20 from B again and 40 from C Ifa customer ordered from C then the next time heshe orders a pizza there is a 25 chance heshe will order from A 50 from B and 25 from C again At the start of the year 30 of the customers order from A 50 from B and 20 from C If customers order pizza once per week find the distribution of customers among the three places after 1 2 and 6 weeks Give the transition matrix Give the initialstage vector Compute the distribution vector after 1 week Compute the distribution vector after 2 weeks Compute the distribution vector after 6 weeks Compute the distribution vector after 1 year Math 166 Lecture Notes 14 Section 14 Introduction to Probability Definition of Sample Spaces and Events Associated with any experiment are outcomes in which we are interested After we decide what to observe we create a list of simple events or simple outcomes ofthe experiment The set of simple events 8 for an experiment is called a sample space for the experiment Simple events are mutually exclusive cannot happen at the same time In each trial of the experiment one and only one simple event should occur We can also de ne other subsets ofthe sample space that represent compound events Example 1 A special die is cast The faces displaying the numbers 1 4 and 6 are red The faces displaying the numbers 2 3 and 5 are blue Various experiments could be defined a Suppose we are interested in the color ofthe upturned face Then the sample space is red blue b Suppose we are interested in the number showing on the upturned face Then the sample space is 1 2 3 4 5 6 c Suppose we are interested in the primecomposite characteristic of the upturned number Then the sample space is prime composite neither d Suppose we are interested in the oddeven characteristic ofthe upturned number Then the sample space is odd even Consider Would red blue odd be an appropriate sample space Whywhy not Probability of a Simple Event in a Uniform Sample Space If S s1s233sn is the sample space for an experiment in which the outcomes are equalylikely then we assign the probabilities Ps Ps2 Ps l to each ofthe simple events sls2s3sn 71 Consider Which ofthe experiments in Example 1 above have a uniform sample space Example 2 Consider an experiment where a coin is tossed and a 3color spinner red blue green is spun We observe the upturned face ofthe coin and the color on which the spinner lands a Give the sample space for the experiment b Assuming this is a uniform sample space find the probability of any event in the sample space any simple event Computing the Probability of any Event in a Uniform Sample Space Let S be a uniform sample space and let E be any event simple or compound Then Number of favorable outcomes in E nE PE Number of possIble outcomes In S nS Example 3 Consider the experiment in Example 2 toss a coin spin a 3color spinner Recall the sample space HR HB HG TR TB TG and that it is a uniform sample space a Find the probability that the spinner will land on red b Find the probability that the coin will land heads up c Find the probability that the spinner will land on red or blue A probability distribution for an experiment is a table of probabilities for each ofthe simple events The table can be in row or column form There should be a row or column for the event descriptions and a row or column for the associated probabilities Example 4 Construct a probability distribution for Example 3 Simple Event Probability Example 5 Suppose we tossthe special die in Example 1 The faces with the numbers 1 4 and 6 are red the faces with the numbers 2 3 and 5 are blue We observe the upturned face a Find Pprime b Find Pcomposite c Find Pneither d Find the probability of getting a numberthat is odd and blue e Find the probability of getting a numberthat is prime and red f Find the probability of getting a numberthat is prime and blue Observations about probability The probability of an event is a number that lies between 0 and 1 The probability of an event that will never happen is 0 The probability ofan event that is certain to happen is 1 The sum of the probabilities ofa simple events in a sample space is equal to 1 Probabilities can be expressed as fractions or decimals When possible probabilities should be given in an exact form Do not convert probability fractions such as 13 to the approximate probability of 333 unless the context clearly implies to do so or unless expressly stated Example 6 Consider a standard card deck A card is selected at random from the deck a Assume each card is equally likely to be drawn Describe the probability distribution b What is the probability that a king is drawn a king or a queen is drawn a heart is drawn a heart is not drawn an oddnumbered card is drawn a black or red card is drawn Example 7 Consider the composition and birth sequence ofa 4chid family all single births a Give examples of three simple events in the sample space b How many simple events are in the sample space Get a mental picture of the tree diagram 0 What is the probability of each simple event in the sample space If the probability ofboy and probability of girl on each birth is equal then this is a uniform sample space Consider it so d What is the probability that a girl will be born first and a boy last e What is the probability that all 4 children are ofthe same sex f What is the probability that there is at least one boy Example 8 Consider again the 4chid family all single births This time suppose we were interested in the probability of3 girls in the family but the birth sequence is not speci cally mentioned a Could we describe the outcomes as S 0 girls 1 girl 2 girls 3 girls 4 girls giving a sample space with 5 simple events and then determine the probability based on that sample space b Why or why not Draw the tree diagram that would be used in Example 7 and Example 8 Use it to construct the sample space Consider How would you like to work it with 10 children 8 c For the 4 children scenario what is the correct probability of 3 girls any order d What is the probability that the 3 girls will be born in sequence Finding the Empirical Probability of an Event If an event E occurs m times in n trials the ratio mn is the relative frequency of the event E after n repetitions Probabilities assigned as a result of experimentation are empirical probabilities This is in contrast to theoretical probabilities assigned as a result of assumptions and deductive reasoning as with dice coins and playing cards Example 9 When tossing a fair coin the theoretical probability is 12 for heads 12 for tails Suppose we conduct an experiment and toss a coin 1000 times and 480 times we get tails and 520 times we get heads a What is the empirical probability that the tossing of that coin will result in heads up Example 10 In orderto help determine insurance rates data was collected about 2000 traffic accidents and the ages and genders ofthe drivers at fault in the accidents The table below shows the results Use the information in the table to finish the probability distribution table and answer the questions below Probability Distribution Age Gender of Age Gender Probability Accidents of Accident Xlt xlt lt 18SXlt25 25SXlt35 lt lt 18SXlt25 25SXlt35 lt lt 35SXlt55 55SXlt65 55SXlt65 X2 2 lt 35SXlt55 55SXlt65 55SXlt65 X2 2 What is the probability that a person who has had an accident where heshe was at fault is a under 25 b at least 25 but under 55 years old 0 55 or older d male e female Math 166 Lecture Notes 43 Section 43 Systems of Linear Equations Unique Solutions The GaussJordan elimination method involves a sequence of operations on a system of linear equations to obtain equivalent systems until a solution is found In linear algebra Gauss Jordan elimination is a version of Gaussian elimination that puts zeros both above and below each pivot element bringing a matrix to reduced row echelon form whereas Gaussian elimination takes it only as far as row echelon form Operations of GaussJordan elimination 1 Interchange any two equations 2 Replace an equation by a nonzero constant multiple of itself 3 Replace an equation by the sum of that equation and a constant multiple of any other equation Example 1 Start with 2X 4y2z 4 3X 8y 2 18 2X y 52 8 Multiply the 1st equation by 3 add the result to the 2nd equation to get a new 2nd equation X 2y z 2 2y 42 12 2X y 52 8 Divide the 2nd equation by 2 X2y2 2 y 22 6 3y3212 Multiply the 2nd equation by 2 add the result to the 1St equation to get a new 1St equation X 52 10 y 22 6 3y 32 12 Divide the 3rd equation by 3 X 52 10 y 22 6 2 2 Multiply the 3rd equation by 5 add the result to the 1St equation to get a new 1St equation X O y 22 2 6 2 Divide the 1St equation by 2 X2y z 2 3X8y z 18 2X y 52 8 Multiply the 15t equation by 2 add the result to the 3rd equation to get a new 3rd equation X 2y z 2 2y 42 12 3y 32 12 Multiply the 2nd equation by 3 add the result to the 3rd equation to get a new 3rd equation X 5210 y 22 6 32 6 Multiply the 3rd equation by 2 add the result to the 2nd equation to get a new 2nd equation It turns out that these operations are a little easier to keep track of if assuming we are careful to maintain the order of the coefficients and constant we strip out the variables and work with the coef cients only in the form of an augmented matrix When we do this the sequence of steps looks like this 24 21 4 38 13 18s 21538 391211 2 1211 2 1211 2 3 8 13 18 H 2 43 12 H 0 2 43 12 H 2 538 21538 0 33312 quot1211 2 10 5110 10 5110 0 1 23 6 s 1 23 6 s 1 23 6 s 0 3 3312 3 3312 0 33 6 quot105110 10010 10010 01 23 6 s 01 23 6 s 0103 2 00132 00132 00132 What we are doing is systematically creating an identity matrix on the left side of the augmented matrix When we are able to do this the solutions are in the column at the far right Your calculator has a function that will do this for you First you will have to enter the values into a matrix in your calculator Press the matrix button 2nd x 1 cursor to the name you want to select A will do Cursor overto EDIT and enter Cursor over to change the dimensions of the matrix rows x columns and enter Enter your values into the matrix It will fill them in across row by row Then you will have to enter the command to create the rowreduced matrix Getting back to the home screen with QUIT 2quot mode is a good idea here Press the matrix button 2nd x 1 cursor overto MATH and cursor down to B rref and enter You should see rref You need to enter the matrix you want it to operate on Press the matrix button 2nd x 1 on NAMES cursor down to your matrix A for us and enter You should find the A has now popped in to your rref function call Close the and enter Now you can simply read off the solutions to your original system of equations Example 2 GaussJordan Eliminationby handa complete example Solve the system using an augmented matrix and GaussJordan Elimination 3x 6y 24 4x3y 10 3 6 I 24 We create the beginning matrix 1 0 I Our final matrix goal is 4 3 I10 0 1 I 39 J Step one Force the 1 in the upper left corner HovW By dividing the entire first row equation by 3 or equivalently multiplying by 13 Symbolically R1 gtR1 3 6 I 24 3 1 2 I 8 becomes 4 3 I10 4 3 I 0 Step two Force a O belowthe 1 just created HovW By multiplying the first row where the 1 is by 4 and adding to the bottom row to create a new bottom row with the O in the desired position However the multiplying of the first row by 4 is done mentally you don t actually change the first row Symbolically 4R1 R2 gt R2 4 I1 2 I 8 Mentally multiplying the top row by 4 I 1 2 weget 4 3 I10 8 and adding to the old bottom row 0 11 I 22I Step three Force the 1 in the lower right corner How By dividing the second row equation by 11 or equivalently multiplying by 111 Symbolically 171le gt R2 1 2 I 8 1 2 I 8 i becomes i 0 11 I 22 11 0 1 I 2 Step four Force a 0 above the 1quotjust created How By multiplying the second row where the 1 is by 2 and adding to the top row to create a new top row with the O in the desired position However the multiplying of the second row by 2 is done mentally you don t actually change the second row Symbolically 2R2 R1 gtR1 1 2 I 8 Mentally multiplying the bottom row by 2 t 1 0 I 4 I we e i 2 I 2 and adding to the old top row 9 1 I 2 Step five Analyze the resulting matrix and determine the solutions to the system Recall that the rows are actually equations The first row translates into the equation 1x0y 4 in other words x 4 The second row translates into the equation 0x1y 2 in other words y 2 The two examples shown above each had a unique solution Not all systems of equations have unique solutions We will have to learn to recoqnize when we have an in nite solutions case or a no solution case For now though using your calculator solve a few systems that have unique solutions Practice these on your own and make sure you understand how to enter the matrices and solve the systems Example 3 Example 4 Example 5 Example 6 Example 7 30X 20y 600 10X 20y 400 X y z1000 80X60y50262800 X y 22400 3OX 10y 202340 10X10y 202180 1OX 30y 202220 O5X 10y 152 380 O6X 09y 122 330 02x 03y 052 120 6w 9X 7y 52250 6w 4X 7y 3z195 4w 5X 3y 22145 4w 3X 8y 22125 Answer X y 10 15 Answer X y z 240 560 200 Answer Answer Answer Xyz824 x y z 20 220 100 w X y z 24 5 2 15 Math 166 Lecture Notes M2 Section M2 Regular Markov Chains One ofthe many applications of Markov chains is in making longrange predictions It is not possible to make longrange predictions with all transition matrices However for a particular group of transition matrices it is always possible If a transition matrix is regular it will always be possible A transition matrix is regular if some power ofthe matrix contains all positive entries no zeroesl A question that will soon come up is this How far must you go with powers ofthe matrix to be certain that a matrix is not regular The answer is that if zeros occur in the same positions in two successive powers of a matrix then they will appear in those positions for all higher powers ofthe matrix Example 1 Is the transition matrix given below regular 25 25 60 T 65 20 10 m T50 10 55 30 Example 2 Is the transition matrix given below regular 2 2 4 T 45 o 6 mT2 1T30 35 80 Example 3 Is the transition matrix given below regular 0 02 06 T 05 08 04 mT2 T3 05 0 0 Example 4 Is the transition matrix given below regular 05 0 04 T0106T2T3 05 0 0 Compute and comment on T 50 Equilibrium vector of a Markov Chain If a Markov chain with transition matrix T is regular then there is a unique vector V such that for any probability vector vand for large values of n T vV If a Markov chain with transition matrix T is regular then there exists a probability vector V such that TV V Vector V is the equilibrium vector of the Markov Chain M The sum of the entries in the probability vector must be 1 Example 5 Find the equilibrium or steadystate vector for the transition matrix 02 01 03 T 06 01 03 02 08 04 Solution Find the probability vector V such that TV V if it exists x 02 01 03 x x LetV y Then 06 01 03 y y Z 02 08 04 2 2 Performing the matrix multiplication we get 2x1y3z x 2x1y3zx 8x1y3z0 6x1y3z y orthe equations 6x1y3zy or 6x 9y3z0 2x8y4z z 2x8y4zz 2x8y 6z0 Add to those the additional equation x y z 1 since x y and z are probabilities for an entire sample space Solve the resulting system 8x1y3z0 6x 9y 3z 0 2x8y 6z 0 x y 2 1 Recall that you can solve the systems of equations using the augmented matrix and rref function in the calculator x 2174 y 3043 z 4783 Example 6 02 01 03 Forthe transition matrix from Example 5 T 06 01 03 02 08 04 Find T5 to 3 decimal places Find T10 How does it compare to T5 Find T20 How does it compare to T5 and T10 Compare the final matrix with your answers from Example 5 Example 7 Recall this earlier statement in the notes If a Markov chain with transition matrix T is regular then there is a unique vector V such that for any probability vector vand for large values of n T vV quot2quot 39o2 01 033920 quot2quot Letv 3 Compute T20v 06 01 03 3 5 o2 08 o4 5 6 3902 01 03 20 639 Letv 1 ComputeT20v 06 01 03 1 3 02 08 04 3 0 02 01 033920 39039 Letv 8 Compute T20v 06 01 03 8 2 o2 08 o4 2 What do you observe Math 166 Lecture Notes 52 Section 52 Multiplication of Matrices Matrix Product The matrix product ofA and B written AB is defined by 31 132 AB a1 a2 a3 an b3 a1b1a2b2 a3b3 anbn b7 Example 1 On a certain day Action Amusements sold 300 children s tickets 100 adult tickets and 50 senior tickets Ifthe price of tickets is 1 50 for children 300 for adults and 200 for seniors nd the total revenue for Action Amusements for that day lfA is a matrix of size m X n and B is a matrix of size n X p then the matrix product ofA and B AB is de ned and is a matrix of size m X p Note 1 Ifthe inner dimensions don t match up there is no product defined Note 2 With matrices multiplication isnotcommutative AB BA Example 2 Find AB 1 1 2 0 3 0 2 A121 andB3012 0 3 4 1 Solution First do a dimension check dimensions ofA X dimensions of B X Does the product AB exist is it de ned If yes what are the dimensions of AB X Second prepare or lay out the answer matriX 1 1 2 0 3 0 2 3 0 1 2 1 2 1 0 3 4 1 Third systematically do the row X column operations Thought question What about BA Example 3 Find the values of k and d in the problem below 3k 2 4 3 0 12 11 4 1dL012JL478J Laws for Matrix Multiplication If the products and sums are de ned for the matrices A B and C then 1 ABC ABC Associative law 2 AB C AB AC Distributive law The identity matrix of size n n rows and n columns is given by 00 10 In 01 1 00 00 The identity matrix has the property that InA A for any n X rmatrix A and B B for any sx n matrix B In particular ifA is a square matrix of size n then IKA A A Matrix Representation Example 4 Show how the following system of linear equations is represented by the matrix equation below it 3X 2y Z 6 3X6y 1 X 3y 4z O 3 2 1 6 x 36 0 y 1 1 3 42 Math 166 Lecture Notes 4 Introduction Sstems of Linear Equations S tems of Eguations 2 x 2 2 equattons tn 2 yanaotes A system of two ttnear equattons tn two yanaotes x and y nas tne generat torm a x y d aft b d etyen two ttnes Lt and Let one and onty one ottnetottowtng tstrue I t Lt and L2 are parattet and otsttnct I 3x 2y 7 X 2w 6 6x 4y 12 4y 10 5x 3 1 21 y 3x 2y 5 9 H 4 Exac gutuntgutmtevsecttun WWW 339 quotD Wm putnts uttntetsecttun uttntetsecttun stgtems of Eguations 3 x 3 3 equattons tn 3 yanaotes A system of tnree ttnear equattons tn tne yanaotes x y and z nas tne generat torm 41 5 a2xbzyczzdz a3xb3yczzd3 Tnts ttme eacn equatton represents a ptane tn 3D Space ratnertnan a ttne tn 2D space Agatn however tnree dtfferertt resutts are posstote Panes P1 Plano P3 5 tntersectat exactty one potnt exactty one sotutton rsect tn 5 ttne or tne enttre ptane tnttntte sotuttons P2 and P3 are parattet and eacn ts otsttnct from tne otners no sotutton F g 1 F m 8 PartesF tt Panes F t Htustrattons of these Cases are found tn most textbooks S tems of Eguations 4 x 4 and larger A ttnear system composed of 4 or more yanaotes tht go beyond tne scope of a geometnc tnterpretatton However tne same posstotttttes occur a exactty one Sotutton o tnftmtesotuttonsy or c no so utton Other note You do not have to have exactly the same number of equations as variables although typical problem sets in beginning algebra courses often do We will see a variety Options for solving 1 You could solve the system by algebraic manipulation using elimination and substitution techniques This is a basic skill you should know but it can be timeconsuming and tedious 2 Use your Tl83 or Tl84 calculator to solve for the point of intersection after you type the equations into Y1 and Y2 3 Use other methods that utilize the power of the Tl83 and Tl84 calculators One such method is GaussJordan Elimination or Gaussian Elimination Application problems of all types can generate systems of equations to be solved Some are linear some are not We are limiting our study to linear systems One such word problem is below Example 1 Mr and Mrs Garcia have a total of 100000 to be invested in stocks bonds and a money market account The stocks have a rate of return of 12 per year while the bonds and the money market account pay 8 and 4 per year respectively They have stipulated that the amount invested in stocks should be equal to the sum of the amount invested in bonds and 3 times the amount invested in the money market account How should the Garcias allocate their resources ifthey require an annual income of 10000 from their investments Solution Assign X y and z to the amount invested in stocks bonds amp the money market respectively Write equations representing the information in the word problem One formula to recall here is I prt interest principle X rate X time Writing the equations from the problem we get and after reformatting Xyz100000 X y z 100000 xy3z Xy3z0 12X 08y 042 10000 12X 08y 042 10000 At this point we use whatever methods are at our disposable andor are allowed to solve the resulting system of equations Example 2 A dietitian in a hospital is to arrange a special diet composed of three basic foods The diet is to include exactly 340 units of calcium 180 units of iron and 220 units of vitamin A The number of units per ounce of each special ingredient for each ofthe foods is indicated in the table Units per ounce Foad A Foad B F0 d Calcium I 0 I 0 Iron I 10 I 0 I Vitamin A 10 o How many ounces of each food must be used to meet the diet requirements C r o 0 o l l l Solution Let X ounces of Food A 340 y ounces of Food B 180 z ounces of Food C 220 Example 3 A manufacturer makes three types of boats 1person 2person and 3 person models and each boat requires the services ofthree departments A 1person boat requires 06 hr in assembling 05 hr in cutting and 02 hr in packaging A 2person boat requires 1 hr in cutting 09 hr in assembling and 03 hr in packaging A 3 person boat requires 12 hr in assembling 05 hr in packaging and 15 hr in cutting The cutting assembling and packaging departments are available a maximum of 380 330 and 120 laborhours per week respectively How many boats of each type must be produced each week for the plant to operate at full capacity Solution Let X of 1 person boats 380 y of 2person boats 330 z of 3person boats 120 Math 166 Lecture Notes M3 Section M3 Absorbing Markov Processes We will now consider a special type of Markov process one in which at least one state is such that when entered cannot be left Such states are called absorbing states A Markov process is called absorbing if the following two conditions are satisfied 1 There is at least one absorbing state 2 It is possible to move from any nonabsorbing state to one of the absorbing states in a finite number of stages The transition matrix for an absorbing Markov process is said to be an absorbing stochastic matrix Example 1 Is the process represented by the transition matrix below an absorbing process Current state 1 2 0 25 0 60 T Next state 1 65 1 10 2 10 0 30 Example 2 Is the transition matrix below an absorbing stochastic matrix Current state 1 0 25 0 60 T Next state 1 65 1 0 2 10 0 40 Example 3 Is the process represented by the transition matrix below an absorbing process Current state 0 1 2 0 0 1 0 T Next state 1 45 0 0 2 55 0 1 Example 4 Is the transition matrix below an absorbing stochastic matrix Current state oo OOAA 0 T Next state 1 2 Example 5 Is the transition matrix below an absorbing stochastic matrix Current state T Next state wNi IO L a o so OO xOA boobw ooooo Example 6 Is the transition matrix below an absorbing stochastic matrix Current state 1 2 3 ooooo o39woo39q Aoooo o xooo bO39wO39wA 0 1 T Next state 2 3 4 More about Absorbing Markov Processes In an absorbing Markov process the longterm probability of going from a nonabsorbing state to an absorbing state is one Also the longterm behavior may be different for different initial states Note that this is different from the situation with regular Markov processes where the longterm behavior is independent of the initial state To see the longterm behavior we begin by reorganizing the states in the matrix so that the absorbing states are rst and the nonabsorbing states next Example 7 Example 5 again Current state Current state 0 2 3 1 2 3 0 1 4 0 1 1 0 0 1 1 0 1 0 1 0 0 1 4 0 T Nextstate becomes new Tquot Nextstate 2 8 0 0 0 2 0 8 0 0 3 1 0 6 0 3 0 1 6 0 And just for referenceit is interesting to note that there is 1 absorbing state and Current state 2 1 1 0 0 0 0 3 0 0 0 0 The longterm probability that everything will end in state 1 equals 1 is certain new T9999 Next state DO 4 DO 00 1 0 2 Example 8 Is the transition matrix below an absorbing stochastic matrix If so rearrange the states so that the absorbing states are first Also observe new T Current state Current state 0 1 2 3 1 3 2 O 0 1 0 0 0 1 1 0 2 5 1 5 1 2 0 3 0 1 8 4 T Nextstate becomes new Tquot Nextstate 2 0 0 0 0 2 0 0 0 0 3 4 0 8 1 0 0 0 0 1 And just for referenceit is interesting to note that there are 2 absorbing states and Current state 1 3 2 O 1 1 0 15 59 99 3 0 1 45 49 new T Next state 2 0 0 0 0 0 0 0 0 quot0quot The limiting matrix above gives us longterm probabilities for this problem The longterm probability that something initially in state 1 will end in state 1 equals 1 The longterm probability that something initially in state 3 will end in state 3 equals 1 The longterm probability that something initially in state 2 will end in state 1 is 15 The longterm probability that something initially in state 2 will end in state 3 is 45 The longterm probability that something initially in state 0 will end in state 1 is 59 The longterm probability that something initially in state 0 will end in state 3 is 49 The longterm probability that anything will end in state 2 or state 0 equals 0 More about Limiting matrices and finding those matrices How could these limiting matrices be calculated without the use ofa powerful calculator that can compute large powers of matrices Or what ifthe calculator sometimes failed in being able to give us exact answers in some cases We will explore theorems that allow us to find the limiting matrix by direct computation First we reorder and partition the transition matrix according to its a absorbing and b I A nonabsorbing states to get am Mb Obxa Bbxb Note a and b are numeric values Then as n becomes large without bound 1m AI 43H Tn heads for the limiting matrix Obxa Obxb Example 9 Example 57 again Current state 1 2 We begin With T Next state 0 1 0 8 1 th IO lt3vo oo xcn I A Reorganized as m Mb Obxa Bbxb 71 AndthelimitingmatrixLIam IquotB 0 bxa 0 bxb o 0 oo o o 1 0 0 01 04 0 Au Br1 m 01 01 0 0 1 01 06 0 5029 2029 0 4029 4529 0 1 1 1 09 04 0 4 p o n 08 1 0 and therefore L Example 10 Example 8 again Current state 1 2 We begin With T Next state Lo39an o 3 0 0 0 1 wwwo booioo ma Aaxb we have bxa Bbxb 1 I 3 Reorganized as 0 2 0 Iam I B1 And the limiting matrix L bxa Obxb Contains the embedded matrix AI B 1 Au Brl 1 iitt EDI t it 15 59 45 49 5 4 and therefore L 0ND Math 166 Lecture Notes L2 Section L2 Truth Tables We will be examining the possibilities for the truth or falseness of logical statements which combine any number of simple logical statements propositions with any of several logical operations When so we will use the basic shown in the table below V We will not be studying these last two in our course In order to examine the possibilities of a compound statement in an organized and logical manner we will build truth tables Important note When we build these tables we will agree to follow a certain pattern for the assignments ofT and F to the various propositions involved lftwo propositions p and q for example are involved we will use this order lfthree propositions p q and r for example are involved we will use this order Based on the number of propositions involved in a logical statement you will build the layout ofthe truth table as shown above The number of additional columns necessary and the statements they represent will vary from problem to problem Each column to be filled in for a truth table will be based on the results ofthe operations A v l a H negation conjunction inclusive disjunction exclusive disjunction conditional biconditional combined with the propositions statements Example 1 Build a truth table forthe logical statement p A q i p v q V V V Notice how in the last column which indicates the possibilities for our logical statement the entries are all True In such a case our logical statement is called a tautology lfthe last column showed all False the statement would be called a contradiction Any other combinations of True and False have no special meaning Example 2 Build a truth table forthe logical statement p v q A q lr r V r V r V V This example is neither a tautology nor a contradiction It is sometimes true and sometimes false Example 3 Build a truth table for the logical statement p A q v r lq r V V V Math 166 Lecture Notes 11 Section 11 Introduction to Sets Set Terminology and Notation A set is a wellde ned collection of objects The objects are called the elements and are usually denoted by lowercase letters a b c the sets themselves are usually denoted by uppercase letters A B C Example 1 O the set of all odd integers is a wellde ned set S the set of all smart people is not a wellde ned set The elements of a set may be displayed using roster notation by listing each element between braces or by setbuilder notation giving a rule that describes the definite property or properties an object X must satisfy to qualify for membership in the set Example 2 Roster notation A a b c z Setbuilder notation A X X is a letter ofthe English alphabet Roster notation B 2 3 5 7 11 13 17 19 23 Setbuilder notation B X X is a prime number To write that the letter b is an element ofthe set A above we would write be A Note that the Greek letter pi is not an element of the set A we would writer e A Two sets A and B are equal ifand only if they have exactly the same elements Example 3 a e i o u a i o e u 12 3 4 5 0 12 3 4 5 If every element ofa set A is also an element ofa set B then we say that A is a subset of B A g B Every set is a subset of itself Example 4 1 4 6 8 g 1 2 3 4 5 6 7 8 a i o g a i o e u a e i o u g a i o e u a e i o is nota subset of a o e u X y lfA and B are sets such that A g B but A at B then A is a proper subset of B A c B A is a proper subset of B ifA g B m there exists at least one element in B that is not in A Example 5 a o c a i o e u and 2 3 5 6 c 1 2 3 4 5 6 7 Note that a e i o c a e i o is FALSE but a e i o g a e i o is TRUE Likewise 235 c 2 3 5 is FALSE but 2 3 5 g 2 3 5 is TRUE For any set A A g A is true however AcA is false We could also say A A The set that contains no elements is called the empty set and is denoted by Q called the null set The empty set is a subset of every set It can also be denoted by Warning It IS NOT denoted by Q Example 6 a List all subsets of the set A m n p How many are there b List all the prayer subsets of the set B 1 2 3 4 How many are there The universal set is the set of all elements of interest in a particular discussion It is the largest in the sense that all sets considered in the discussion of the problem are subsets of the universal se Example 7 Give the smallest possible universal set for each situation a Determine the ratio of female to male students in Portland College b Determine the ratio of female to male students in the business department of the Portland College Venn diagrams are used as a visual representation of sets The universal set U is represented by a rectangle and subsets of U are represented by regions lying inside the rectangle Example 8 Use Venn diagrams to illustrate the following statements a AB b ACB c AqBandBczA Set Operations The union of sets A and B A u B The intersection of sets A and B A m B is the set of all elements that belong is the set of all elements to either A or B or both in common with the sets A and B AUBXXEAOFXEBOI39bOth AmBxXEAandXEB U 7i A A Two sets are disjoint if they have no elements in common that is ifA m B Q Exagle 9 Let A v w x y and B x z w U Illustrate with a Venn diagram A Find A u B Find A n B Exaggle 1o lfA 1 3 5 7 9 and B 2 4 e a 10 U Illustrate with a Venn diagram Find A u B A E Find A n B in this case We would say that sets A and B are Exaggle11Lquot39 39 I LetA 10 Iu quot r 39 Iu Illustrate with a Venn diagram U Him First give a roster listing of the sets B U i Bi If Uis a universal set and A is a subset of U then the set of all elements in Uthat are not in A is called the corrplement ofA and is denoted A A x x e U x e A Exa le 12 Let U1 2 3 4 5 6 7 B 9 10 A 2 4 6 810 and B 12 4 5 7 10 Drawa Venn diagram Find A Find 8 If Uis a universal set and A is a subset of U then a U z b 21 c A A d AUACU e AnAcZ Example 13 Usmg a Venn dtagram snade a A uBnC o AUBYMC A A v v A 5 Let U be a untversat set tt At EL and C are arottraty subsets of U tnen Commutauve aw tor unton Commutauve aw tor tntersectton B c C Assoctattve taw tor unton s tattve taw tor tntersectton C Dtstrtouttve text or n C Dtstrtouttve aw tor tntersectton De Morgan s Laws AU3 A n3 d Ans A U3 U 6 ntsn gammy magma Exam le14 Ustng Venn dtagrarnst snowtnat was ACUBC U U Examgle 15 Let U m 2 3t 4t 5t 6 7t 8 9 to A m 2 4t 8 91 and B 3t 4t 5t 6 8 snow by dtrect cornputatton thatAu8 14 3 Au BC AC n BC Example 16 Let Udenote tne set ot aH students currenuy enroHed at AampM and x e Ut U x e Ut x ts taxtng a oustness ctass c x e Ut x ts taxtng a computer dass Fwd an expresston tn terms ot At 5 and ctor m Tne set ot AampM students taxtng an art and a 39 oustness ctasst out not a computer ctass o Tne set ot AampM students taxtng a computer ctass out not a oustness or an art ctass Example 16I continued Let U denote the set of all students currently enrolled at AampM and A X e U X is taking an art class B X e U X is taking a business class C X e U X is taking a computer class Draw the Venn diagram and describe each in words U U A V 6 AUG0 A BUC 9quot OPquot 3 Describe the area the instructor shades in class Describe the area the instructor shades in class OPquot 3 Describe the area the instructor shades in class Describe the area the instructor shades in class Example 17 Let U 13 5 7 9 11 13 15 17 19 21 LetA3 5 7 9 15 LetB7 91 15 19 LetC 137111517 Note This may be done by direct computation or by use of a Venn diagram Find the sets described in a b and 0 below using direct computation a BUCC b BUAnC C BcnAnCc Find the sets in a b and 0 above using a Venn diagram U 6 U13 5 7 9 11 13 15 17 19 21 A3 5 7 9 15 B79131519 c137111517 a BUCC b BUAnC C BcnAnCc Check your answers on the following page Example 17 answers Let U 13 5 7 9 11 13 15 17 19 21 LetA3 5 7 9 15 Let B79 13 15 19 LetC 1 3 7 11 15 17 Find a BUCC13791113151719C 521 b BUAnC BU 3 7 157 9 13 15 19 u 3 7 15 3 7 9 13 15 19 c BcnAnCc Bcn3 5 7 9 15n 5 9 13 19 21 Bcn5 91 3 5 11 17 21n 5 95 4 Math 166 Lecture Notes 33 Section 33 Measures of Spread Variance and standard Deviation Two groups of data can be simiiar in that they couid have the same average mean expected venue and yet the data sets couid aiso be quite ditterent One possibie i whereas in the other data group the data might be more cioseiy ciustered near the mean ance The variance of a random variabie associated with the probabiiity distribution is a way of measuring this degree of dispersion or spread Detinition Suppose a random variabie has expected vaiue EX u x x x and the probabiiity distribution 2 3 quot x PX 2 P Then the variance of the random variabie X is given by VIM pm T P X2 02 P7 02 Note The variance of the random variabie x is measured in square units p st ndard Deviation T standar dev ation of a random variabie X denoted with 0 is aiso a way of measuring the degree of dispersion or spread Detinition Suppose a random variabie has expected vaiue 5m u and the probabiiity distribution shown above Then the standard deviation of the random variabie x is given by ks mm 4 px72pzxz z PXT2i N ole Example 1 The two histograms beiow have the same expected vaiue u M5 u i v Figure 1 Figure 2 VarX223 5149 VaTX413 5203 Population Statistics vs Sample Statistics When we have data for every single object in a universal set then we call the statistics calculated on this data set quotpopulation statisticsquot For example if we are interested in finding the average height of 18yearold males in the state of Texas our universal set would be ALL 18yearold Texas boys lfwe could get the height of every single one of these 18yearold boys then the mean standard deviation and variance would actually be the population mean population standard deviation and population variance Since in many cases such as this it is nearly impossible to get data on EVERY single element of the universal set we take a random sample ofthis set and collect the data on the sample The idea is that as long as our sample is taken randomly the sample39s mean standard deviation and variance quotsample statisticsquot should be good estimates of the actual population statistics The notation for population and sample statistics is different Population mean u or EX Sample mean 9 xbar Population standard deviation 0 Sample standard deviation 8 Population variance 72 or Var X Sample variance 82 Computationally population mean and sample mean are the same However population variance and sample variance are calculated slightly differently for speci c reasons beyond the scope of this course Therefore when computing standard deviations and variances one should acknowledge the difference between a population when we have data on every single element in the universal set and a sample when we have data on only a subset of the universal set Using the builtin calculation functions to find variance and standard deviation If you are working from a list of data enter the data into a stat list and follow the calculator operations below To calculate the mean variance and standard deviation for a single list of data STAT and choose 1 Edit Enter the data in a list eg List 1 STAT then right arrow to CALC and choose 1 1Var Stats Press the 2nd function button and identify the appropriate list eg L1 for List 1 On your home screen you should see 1Var Stats L1 Press Enter On the resulting screen you can find the mean and standard deviation To now calculate the variance VARS and choose 5 Statistics then 4 ex standard deviation press x2 enter Or simply enter the standard deviation with plenty of decimals and square it If you are working from a frequency table or probability distribution enter the data into two separate stat lists and follow the calculator operations below To calculate the mean median amp standard deviation when given the probabilities or frequencies Put the data in L1 and eitherthe probabilities or the frequencies in L2 Follow the same steps including the 11Var Stats option However when you identify the list identify both lists separated by a comma On your home screen you will see 1Var Stats L1 L2 Example 2 The number of candies in ten bags each of brand A and brand B gummy worms were counted The results are shown in the table LetXand Ydenote the random variables whose values are the number of candies in the bags of brand A and brand B gummy worms respectively Use your calculator to compute the standard deviations and variances ofXand Y Numbers of Candies IBrandA 24222o2322212o242223 BrandB 23232221232322212222 Standard Deviation Variance BrandA X Brand B Y Example 3 Once again use your calculatorto compute the mean standard deviation and variance ofX Brand A This time use the probability distribution below Also give the mean median and mode Brand AX Number of candies 20 21 22 23 24 IPX I Probability I 02 I 01 I 03 I 02 I 02 I I Brand B Y I Number of candies I 21 I 22 I 23 I IPY I Probability I 02 I 04 I 04 I Mode Median Mean Standard Deviation Variance BrandA X Brand B Y Variance and Standard Deviation in a Binomial Experiment For binomial experiments with the earlier de nitions of X n p and q the mean expected value variance and standard deviation ofXare H EX quot3 VarX quotW oquot vnpq Example 4 Once again recall the clothes dryers problem 78 ofthe dryers were good 22 needed repair Cindy purchased 20 ofthem for her laundromat Compute the variance and standard deviation for this problem Remember the variance and standard deviation are based on the random variable values and the associated probabilities We answered several questions about Cindy s dryers but we didn t construct a probability distribution It would have had 21 entries for the possible number of defective dryers Without having to construct the entire table and then enter the table values into the calculator we can use the formulas above Variance Standard Deviation Math 166 Lecture Notes 32 Section 32 Measures of Central Tendency Mean Median Mode Expected Value Mean average The mean of the n numbers x1x2xn is E read X bar and is e x x x gIven by the formula x 71 Median The median is the middle number ofa set of data when the numbers are arranged in order of size Ifthere is no single middle because the number ofentries in the data set is even then the middle two numbers are averaged Mode The mode ofa set of numbers is the one that appears more frequently than the others Iftwo numbers appear the same number oftimes so that there is a tie forthe mode then we say the set is bimodal and has two modes Ifthere is no such number or two numbers then the set has no mode Example 1 An observer at Freebirds on a certain day recorded the number of customers waiting in line at the beginning of 5minute intervals between 11 am and 1 pm What was the average number of customers waiting in line on that day Suppose the recorder wrote down this data as the observation was taking place 2123355443556565434442201 Mean To find the mean average we add up all 25 numbers and divide by 25 The mean is 352 Median To nd the median we must line the numbers up in order O112222333344444455555566 and find the middle number ortwo midde numbers and average them The median is 4 Mode To nd the mode we must tally each number or order and countsomething This set is bimodal 4 amp 5 occur the greatest number oftimes and occur six times each If someone asked If you go to Freebirds between 11 am and 1 pm how many people are usually standing in line Which would be the best answer 352 4 or 4to 5 or The truth of the matter is that ofthese three measures of central tendency sometimes one is a better reflection ofthe data than another 80 reporting all three can sometimes be quite helpful in case one or more is skewed Later we will discover some other measures that also help If we organize the data into a frequency table such as the one below we can benefit from a more compact and more easily readable form ofthe data Fre uenc table of Customers waiting in line O FrequencyofOccurrence 1 2 3 4 5 6 2 4 4 6 6 2 Also finding the mean or average is quite convenient when using this format We multiply each data number times its frequency and then add and finally divide by 25 We get 01 12 24 34 46 56 62 88 and 8825 352 Instead of adding up and then dividing what if we divide each piece by 25 and then add Well if we do that we end up computing what we call Expected Value which is essentially the same as the average and mean The difference is that the expected value is normally generated from a probability distribution Probability Distribution of Customers waiting in line 0 1 2 3 4 5 6 Probability of Occurrence fraction i l i i i i l 25 25 25 25 25 25 25 EX 0125 1225 2425 3425 462 5 5625 6225 3 1325 352 Of course it doesn t matter if we use the probabilities in fraction form or in decimal form Probabilit Distribution of Customers waiting in line O Probab yofOccunencededmaD 04 EX 004 108 216 31 Expected Value Let Xdenote a random variable that assumes the values x x 123456 08 l 16 l 16 l 24 l 24 l 08 6424524608352 x with associated 1 2 n probabilities plp2pn respectively Then the expected value ofX EX is given by EXx1p1 x2p2xnpn Note The numbers x x 1 2xn may be positive a profit zero or negative a loss Let s revisit the median and mode from the probability distribution Probabilit Distribution of Customers waiting in line I O I 1 2 3 4 5 6 Probab yofOccunencededmaD 04 l 08 l 16 l 16 l 24 l 24 l 08 l Using the table can you easily determine the median and the mode Compare back to the frequency table if necessary Mode Median ExamEIeZ L t tn ta tnatran opperrnost wnen W0 tarr drce are cast Fwd tne expected xatoe 500 of X X5umofdtce PROBABILITY x Pgtlt E00 2 3 4X 5 6X 7 8 9 10 M 12 7 0t r drawrng tor pnzes at tne end of tne contest Tnere M be a grand pnze worth 50 tnree 2 pnzeswortn 25 eacn andfwe 339 pnzeswortn 10 eacn Kent rs rnore concerned ExamEIeS My order to rarse moneyfor chartty a contest rs netd ata tocat oosrness wnen an ne cnanty prontrngtrorn tne event mums oottn at 500 w srngte donatronswere rnade tr Kent was one of tnose who rnade a 5 donatton compote nrs expected pront contest Expected Vatoe Egtlt Caution f someone tnesto teacn you a shortcut to gettnrs answer drrrerentty t t may sxrps stepst rnost see and 2 t may not Worx on 3 prooternsm Example 4 David s mom creates a game On Fridays she gives David his allowance of 10 Then David spins a 10sectioned spinner with three blue sections two red sections and five green sections If it lands on red his mom gives him five extra dollars If it lands on blue she gives him three dollars If it lands on green she gives him two dollars David must put all the money in his savings account Afterthe 10th week of playing this game how much should David have in his savings account No other money is in the account other than the money from the game Solution Let X represent the different amounts David may place into his savings account at the end of each week Fill in the table Compute the expected value and explain it I Event Outcome I I I I X Possible de osits I I I I Probabilit EX After 10 weeks Example 5 David s mom modi es the game Her goal is for David to put an average of 15 each week into savings This time if the spinner lands on blue she gives him 10 extra and if it lands on green she gives him 2 extra n orderto achieve her goal what is the amount she should give David if the spinner lands on red m I Event Outcome I I I I X Possible de osits I I I I Probabilit EX EX Amount for red Note Sometimes in a gaming situation a problem like the one above may want you to find the value of a particular outcome if you want the game to be fair In other words EX 0 Example 6 The salaries of 12 executives in a company were as follows 130 K 125K 110K 80K 90 95K 105K 360K 80K 80K 95K 270K Find the mean median and mode ofthese salaries Mean Median Mode Which measure do you think is most representative ofthe data If you were hiring a promising young recruit which might you use to tempt himher If you were applying for work in this company which would you want to be told Example 7 Two hair salons operate in the same city Salon A has 20 appointment times available each day Salon B has 25 appointment times available each day Data was kept daily forthe last two months recording the percentage of the daily appointment slots that were filled each day The results are given in the tables below The average profit per appointment at Salon A is 6 and the average profit per appointment at Salon B is 5 Saon appt slots filled X 040 050 060 070 080 090 A Probability 010 010 015 040 020 005 Saon appt slots filled Y 020 030 050 060 070 080 100 B Probability 010 015 015 020 020 015 005 a Find the average number of appointment times filled per day at each salon Hint first find the average of appt slots filled then the average of slots filled Salon A EX Average of appointments filled each day Average of appointments per day X Salon B EY Average of appointments filled each day Average of appointments per day X b Which salon has the highest average daily profit Salon A daily average pro t X Salon B daily average pro t X MeanI or Expected ValueI of Binomial Distributions For binomial eXperiments with the earlier de nitions ofX n and p we use ufor the mean or eXpected value p EX np Example 8 Remember the clothes dryers problem 78 of the dryers were good 22 needed repair Cindy purchased 20 ofthem for her laundromat fthe factory always ships out dryers in lots of 20 what is the average number of machines per lot that will need repair before the warranty is up Math 166 Lecture Notes 21 M Section 21 M Distinquishable Permutations involving Indistinquishable Objects Example 1 Find the number of permutations that can be formed from all the letters in the word CABINET Solution All the letters in the word CABINET are different From 7 letters we are choosing all 7 to make an arrangement The number of permutations is P77 5040 Example 1 was simple since all the letters in the word cabinet are different It gets more complicated Example 2 Find the number of distinguishable permutations that can be formed from all the letters in the word ATLANTA Discussion Example 2 The idea here is that for a given permutation such as ATATLAN rearranging the two T s or any of the three A s doesn t give another unique arrangement If your back was turned when the T s were switched you wouldn t know the difference The number generated by P77 would include ATATLAN and ATATLAN I switched the T s can t you tell Since we don t want those types of switches counted as different arrangements we have to divide out the duplicates Permutations of n Objects Not All Distinct Given a set of n objects in which n1 objects are alike and ofone kind n2 objects are alike and ofanother kind and finally n objects are alike and of yet another kind so that n n2 n n then the number of permutations of these n objects taken n at a time is given by n nllnzln l Solution Example 2 The computation L or will divide out the duplication in 23 23 the 2 T s and the 3 A s There are 420 distinct permutations Example 3 There are 6 black dogs 3 brown dogs and 2 white dogs to be lined up for a dog show How many different color sequences are possible Solution We would generally assume dogs to be unique so if it was a question of simply ordering the dogs forthe show the answer would be P11 11 However if only the color sequences are under consideration then the 6 black dogs are all identical in that respect We need to divide out the duplicate sequences that would be formed by switching for example any 2 black dogs 11 P1111 or 632 632 sequences There are 4620 ways unique color sequences The computation will divide out the duplication in the color Math 166 Lecture Notes 16 Section 16 Conditional Probability and Independent Events Conditional Probability The probability of an event occurring given that another event has already occurred is a conditional probability Example 1 involves conditional probabilities However the situation is simple enough to Number of favorable outcomes in E nE stIII compute usmg the formula PE 7 Number of possrble outcomes In S nS Example 1 Two cards are drawn without replacement from a wellshuffled deck of 52 playing cards a v What is the probability that the rst card drawn is a diamond b v What is the probability that the second card drawn is a diamond given that the rst card drawn was a diamond C v What is the probability that the second card drawn is a diamond given that the rst card drawn was not a diamond As problems get more complicated we will need to resort to a different formula Conditional Probability of an Event IfA and B are events in an experiment and PA 3 0 then the conditional probability that the event B will occur given that the event A has already occurred is PB A PAnB or PO 14 Number of elements In AnB nAnB PA Number of elements in A nA Example 2 A pair of fair dice is cast What is the probability that the sum ofthe numbers o o falling uppermost is 8 if it is known that one of the 39 H 390 quot 05 3 numbers is a 3 o 11 12 13 14 15 15 PBIA quot1403 39 21 22 23 24 25 26 31 3 2 3 3 3 4 3 5 3 6 8n1dieisa3 39 39 Psum81dieisa3 n1dlelsa3 41 42 43 44 45 45 51 5 2 5 3 54 5 5 5 6 61 6 2 6 3 64 6 5 6 6 PAnB Observe the alternative use of PB A PA PAn 5 Ushg afgebrafc rhahfpmatfoh we can changetheformman A PM We the Product Rule PAnB PA PB A ExamEIeS There are 200 ahfrhafs W Seuss Zoos ofwmch 120 are rharhrhafs Amer 70 of the rharhrhafs and 60 of the Ho rmamma have been at them for more than 3years f 8 Trytms P3 nM P3PM 3 Thehthfs PMn3quotPMP3 r M b the ahfrhaf fs hot a rharhrhaf and has been atthezoo 3 years or ess PM5 ns PM5P3 M5 Tree Diagrams Example 4 Draw a tree dfagrarh to fhustrate tosses O r COW Assfgh probabmues to each branch I E 4 I I Are the probabmues ch the 2 set of branches affected by the prewous outcome d Example 5 Draw a tree dfagrarh to fhustrate 2 draws wfthout repfacerheht from a be R c g contafnm91 red and 3 green rharbfes a Are the probabfhtfes on the 2 set of branches affected by the prewous outcome7 c M M c Example 6 tmagtn 52 cards wnat woutd tt took We Cententh nete Wrtn card cecA questtbns we can took at tne questtbns betng asAec and bsbatty de ne our brancnes based on tne questtbns dramond vs not a crarnbnc face card vs not a face card etc and not have to use 52 dtfferent brancnes Tree Dragrams for Condrtronat Probab tres Armite stochastic process ts an expenrnent ConStSUng of anntte number of stages n eacn and assoctated probabtttttes of me precedth stages vanety of probabnmes the Escn set Ufbrancnes represents a stage tn tne brbcess Example 7 Use tne tree dtagrarn to nnd tne tndtcated probabnme a A as E b WE A DA A U7 c WARE netE d PE us me E a 2 9 WE nutE SUM to come PtAtE Example 8 At a tocat CoHege tt Was found that of the 1000 new Students 400 mate and 600 fema e 40 of the men and 20 of the Women Were an majors Gwen that a new major A PW M MAO M or PAn M MM PM M not A A Observe tne retattensntps between tne dtagram and tne termuta not A Example 9 Pam s W0 CoHector S buffa o mcke s Were acadean dropped m a bank wtth ten other mcke s Pam shakes ohe mcke out ata Ume ngto two hts outtato mcke s outtato mcke out after t a one N E E o two thes ME E nutE c three mes huta Example 10 Two cards are drawn wthout reotaoemeht from a deck ot 52 otaythg cards 8 oectoe howto draw the otagram Asstgh prooaothttes to each branch b V A h gweh that the hrst card orawh was a otamoho c Fmdthe probabmtymatme second card orawh ts a otamoho d a otamoho e Fmdthe prooaothtythatthehrst card ts hota otamoho t eweh that the hrst card orawh was not a dtamond tho the prooaothty that the second card orawh was atso hota otamoho Independent Events In general two events A and B are independent if the outcome ofone does not affect the outcome ofthe other lfA and B are independent events then PAl B PA and PBl A PB Test for the Independence of Two Events Two events A and B are independent if and only if PAnB PAPB Note Do not confuse independentevents with mutually exclusive events Independence means the occurrence of one event does not affect the occurrence of the other event Mutualy exclusive events cannot occur at the same time Independence of More Than Two Events If EE2En are independent events then PE1 er2 nunEn PEPE2PEn Note The converse of the theorem is not necessarily true that is knowing that PE1 nEZ n nEn PE1PEZPEK does NOT guarantee that the events are independent Example 11 Consider the experiment consisting of spinning a fairtricolor spinner red blue green twice and observing the outcomes Show that the event A blue in the first spin and the event B red in the second spin are independent events Solution It seems that for this problem the simplest approach is to start by defining the sets 8 A B and A08 8 A B mm Then determine the probabilities PA nB PA PB Then apply the test doesPA m3 PA PB Example 12 A survey conducted by a math professor found that of 400 Math students 50 made an A on the rst exam and 150 studied 6 or more hours per week Ofthose who made an A forty studied 6 or more hours per week Using this data determine whether the events A making an A on the exam and Bstudying 6 plus hours weekly are independent events Solution We could draw a tree diagram label the probabilities and use it to answerthe question However this is a case where it is much simplerto use the formula PE nEnS Determine the probabilities PA nB PA PB Then apply the test doesPA m3 PA PB Example 13 It is known that the three events A B and C are independent and PA 08 PB 04 and PC 05 Compute a PAnB b PAanC Example 14 Billy has invented a gadget to help moms with their housework It has three major components a motor a gear and a battery Billy has determined that on the average 2 of the motors 1 ofthe gears and 3 ofthe batteries are defective Assume that the defects in the manufacturing ofthe components are unrelated a Determine the probability that a gadget selected at random coming off Billy s assembly line is defective b Determine the probability that a gadget selected at random coming off Billy s assembly line is not defective Question What is the relationship between the probabilities described in a and b above Analysis question What would constitute a defective gadget a nondefective gadget A defective gadget would be one where any one of the components is defective A gadget that is notdefective then would be one where the motor is notdefective AND the gear is notdefective AND the battery is not defective Construct a tree diagram for this problem Let M be the event that the motor is defective G be the event that the gear is defective and B be the event that the battery is defective Using the tree diagram determine the answers to a and b above Which one is the simplest to compute Pdefective or Pnot defective Find that one How can you use that to nd the other one Math 166 Lecture Notes 17 Section 17 More Conditional Pr abilites A Posteriori Probabilities ob iously discussed tne cond i PtslA As for as tne probability computations tne latter is represented by a single section of a brancn on tne tree diagram niletne otner inyolyes a more complicated l manyt xtbooks a ratnertorebodingtormula called Bayes Theorem is computation e presented to compute tne a posteriori probabiliti HMPVPI39 bability by using tne formulas 8 Win eitner case We could find tne conditional pro B P AM An i and Nate i Examplel Usingtnetree diagram below andtne formulas shown above compute Em and Mia B 05 8 WSW D PAlB ol a quotMA 397 C Example 2 snops A 5 and 0 supply sonu sonu and 20 respectiyely of tne cabinets used man u i 0 of tne cabinets tnat were produced by snop A are oak and 70 of tne cabinets p below a eiyen tnat a randomly selected cabinet in inyentoiy is trom Company By wnat is tne u o 7 probability tnat it is oak D A n s quotm 0 n 7 O U 3 B lt An oakcablnetls randomly selected Wnat is tne yK not 0 H B7 C El 7 O 3 probability tnat it came from s op not 0 le3 ah u area hrgh schoots 30 are graduates of Adams Htgh 50 are graduates of Behwew Htgh and 20 are graduates of other schoots Statrstrcs show thatthe graduates trorh Adarhs BeHvteW and the other schoots earh therr assocrate degrees trorh Cardone at rates of 00 70 and 20 respectwew MN 8 Htgh b was a graduate of Behwew Hrgh ahd recewed a degree trorh Cardone c t one 0 Cardone WM recewe a degree from Cardone d was a graduate of Beuwew Htgh 9 what rs the probabmty that heshe wru recewe a degree trorh Cardone t Behwew Hrgh or wru comptete therr studres at Cardone wrth a degree Examgle 4 who planned to attend college Tne table below glves tne results ot tnls Study SAT scores Students in that Students in score range score range attending college 1500 71000 5 90 1400 71499 10 80 1200 71399 30 70 1000 a 1199 25 50 800 a 999 20 30 Under 800 10 10 meng et lees a panel tree magmm nldn nelp ln ermrlndtne dueslms a wnat ls tne prooaolllty tnat a Student selected at randorntrorn Jackson ngn nas a SAT score ot1500 or greater and ls plannlng to attend college 9 lt a randomly cnosen Studentfrom Jackson ngn ls plannlng to go to college wnat ls tne prooaolllty tnat tne Student nas an SAT score ot 1200 to 13997 c lt a randomly cnosen Studentfrom Jackson ngn ls plannlng to go to college wnat ls tne prooaolllty tnat tne Student nas an SAT score ot under 10007 Math 166 Lecture Notes 21 Section 21 The Multiplication Principle and Permutations The Fundamental Principle of Counting The Multiplication Principle Suppose there are m ways of performing a task T1 and n ways of performing a task T2 Then there are mn ways of performing the task T1 followed by the task T2 Note This principle can be extended to any number oftasks T1 through Tk Explore the problems below by drawinq a partial tree diaqram and also by appvinq the multiplication principle Example 1 Margaret has decided to purchase a dining room table a set of chairs and a hutch She has narrowed her choices down to three different tables two styles of chairs and two hutches In how many ways can she make her selections Example 2 At the Ice Cream Palace one can choose from two kinds of cones or a cup three choices of ice cream avors and two kinds of toppings How many different ways are there to create a dessert that consists of a single flavor scoop and a single topping Example 3 A coin is tossed and then a threecolor spinner is spun and then a 4sided die is cast The results are recorded Compute the number of possible outcomes ofthis activity Find the answers to the problems below by applyinq the multiplication principle Picture a tree diaqram in your head ifithelps Example 4 A six character password must be created that begins with a letter ofthe English alphabet followed by 5 numeric digits Determine the number of possible passwords if no repetition of digits is allowed Example 5 License plates are created by using two letters English alphabet uppercase followed by four digits lfthere are no restrictions on repetition how many different license plates are possible Example 6 In registering for a College Algebra class at a community college a student has a choice of three class formats computeraided instruction lecture or online Each format is offered from each of the four campus locations Each format of class can be chosen either in a required calculator format or a noncalculator format How many different options are possible for a student signing up for College Algebra Example 7 A family planning to have five children are curious about the number of sequences that are possible forthe sexes of the children Assuming no multiple births how many possible sequences are there Example 8 Two main roads and three back roads connect towns A and B and one main road and two back roads connect towns B and C Use the multiplication principle to find the number of ways a journey from town A to town C via town B may be completed Example 9 Tom s garden will have ten rows of plants The middle two rows will be com One outside row will be yellow squash and the other outside row will be zucchini squash Of the six remaining rows there will be one row each of peas beans okra tomatoes beets and onions lfthere are no additional restrictions how many different garden layouts are possible Example 10 A quiz consists of ve TF questions and five multiple choice questions with four choices each In how many different ways can the test questions be answered Example 11 How many even 4 digit numbers exist between 5000 and 9000 inclusive Introduction and Calculator Notes for Factorials and Permutations Nfactorial a useful definition for the material in this section For any natural number n nfactorial is given by nnn 1n 2321 O Factorials to compute n using your calculator 1 Enter n on your home screen 2 Press MATH cursor overto PRB and down to 4 3 Press ENTER Example 12 Verify the equations below using the function in your calculator a 5 54321 120 b 8 87654321 40320 Permutations A permutation of a set of distinct objects is an arrangement ofthe objects in a de nite order Note laterwe discuss 39 39 quot where order is irrelevant The number of permutations of n distinct objects taken n at a time Pn n n n rt r The number of permutations of n distinct objects taken rat a time Pn r Permutations to compute nPr sometimes written Pn r 1 Enter n on your home screen 2 Press MA TH cursor overto PRB and down to 2 nPr 3 Enter r 4 Press ENTER Example 13 Verify the equations below using the formula a P6 3 120 b P10 2 90 Example 14 Verify the equations below using the function in your calculator a P9 4 3024 b P5 5 120 Example 15 Let A X y z a How many permutations are there of the set A Answer P3 3 or 3 6 List the permutations ofA Answer X y z X z y y X z y z X z X y z y X C39 Example 16 Find the number of ways a baseball team consisting of nine people can arrange themselves in a line for a group picture Solution The standing order is part of what makes one picture different from another We are dealing with a permutation We are choosing from a group of 9 objects choosing all 9 There are P99 or 9 362880 ways the nine people can arrange for the picture Example 17 Find the number of ways a representative group of3 from the nineman baseball team can arrange themselves in a line for a picture Solution The arrangement matters so we will use permutations From a group of9 we are choosing 3 There are P93 504 ways to select a subgroup of 3 people from the team and arrange them for a picture Example 18 Find the number of ways the twelve men in a garden club can choose a President VicePresident and SecretaryTreasurer Solution The arrangement matters so we will use permutations From a group of 12 we are choosing 3 There are P123 1320 different ways to choose a President VicePresident and SecretaryTreasurer Example 19 Nan has 4 history books 5 grammar books and 7 math books If Nan is going to place the books on a shelf where only three of each type can be displayed and all the books of each type are together how many arrangements are possible Solution Task 1 Select and arrange the history books P4 3 24 Task 2 Select and arrange the grammar books P5 3 60 Task 3 Select and arrange the math books P7 3 210 Task 4 Arrange the groups P3 3 6 Total number of arrangements 24 x 60 x 210 x 6 1814400 Note This problem uses both the concept of permutations and the concept ofthe multiplication principle Example 20 Find the number of permutations that can be formed from all the letters in the word CABINET Solution All the letters in the word CABINET are different From 7 letters we are choosing all 7 to make an arrangement The number of permutations is P77 5040 Note It is important to realize that the letters in the word are all different fthey were not the solution would be slightly more complicated Distinquishable Permutations involving Indistinquishable Objects Example 21 Find the number of visibly different arrangements distinguishable permutations that can be formed from all the letters in the word ATLANTA Discussion Example 21 The idea here is that for a given permutation such as ATATLAN rearranging the two T s or any of the three A s doesn t give another visibly unique arrangement If your back was turned when the T s were switched you wouldn t know the difference The number generated by P77 would include ATATLAN and ATATLAN I switched the T s can you tell Since we don t want those types of switches counted as different arrangements we have to divide out the duplicates Permutations of n Objects Not All Distinct Given a set ofn objects in which n objects are alike and ofone kind n2 objects are alike and ofanother kind and nally n objects are alike and of yet another kind so that n1 n2 n n then the number of permutations of these n objects taken n at a time n is given by nllnzln l Solution Example 21 The computation L or 239339 23 the 2 T s and the 3 A s There are 420 distinct permutations will divide out the duplication in Example 22 There are 6 black dogs 3 brown dogs and 2 white dogs to be lined up for a dog show How many different color sequences are possible Solution We would generally assume dogs to be unique so if it was a question of simply ordering the dogs forthe show the answer would be P11 11 However if only the color sequences are under consideration then the 6 black dogs are all identical in that respect We need to divide out the duplicate sequences that would be formed by switching for example any 2 black dogs l 11 or P1111 632 632 There are 4620 ways different color sequences The computation will divide out the duplication in the color sequences Example 23 Jim is organizing his closet He will hang his slacks together in groups by color He has 6 black slacks 3 of which are identical He has 4 pair of grey slacks 2 of which are identical He has 5 pair of khaki slacks all of which are identical In how many distinguishable ways can he order the slacks in his closet P66 P44 P55 3 X 2 X 5 Solution P33 X Finish the work above for a final count Can you explain the parts in the solutions Math 166 Lecture Notes 13 Section 13 Experiments Sample Spaces and Events Terminology An experiment is an activity with observable results The results ofthe experiment are called the outcomes ofthe experiment The set consisting of all possible sample outcomes or sample points of an experiment is the sample space Each repetition of an experiment is called a trial Tree diagrams can be used to diagram the trials and the sample space of an experiment A sample space associated with an experiment that has a nite number of possible outcomes is called a finite sample space Sample spaces can also be in nite A subset ofa sample space of an experiment is an event The empty set represents an impossible event The sample space itself represents a certain event Two events E and F are mutually exclusive if E n F Q they can t happen at the same time Venn diagrams can be used to depict the union intersection and complementation of events Example 1 Experiment Toss a coin and observe whether it falls heads or tails a Describe the sample space associated with the experiment 8 H T is the sample space This is the listing of all possible outcomes for a single coin toss b What are the events of this experiment all subsets of S H T H T Note H T is the certain event is the impossible event Example 2 Experiment Cast a 6 sided die Observe the numberthat falls uppermost a Describe the sample space associated with the experiment 8 1 2 3 4 5 6 is the sample space ofthe experiment This is the listing of all possible outcomes of one toss ofa die b List two or more events of this experiment interpret the meaning E 2 4 F 1 2 5 and G 1 6 A few subsets ofthe sample space E is the event that a 2 or 4 falls uppermost F is the event that a 1 or 2 or 5 falls uppermost What is G 0 Compute E u F E n F E n G and Fc interpret E U F E n F E n G Comment FC Example 3 The organizer of a benefit concert records the number of patrons attending a b C The theater has a seating capacity of 200 What is an appropriate sample space for this experiment Describe the event E that more than 150 people attend the concert Describe the event Fthat the theater is less than half full at the concert F Example 4 An experiment consists oftossing a coin three times and observing the a b C resulting sequence of heads and tails Determine the sample space S ofthe experiment A tree diagram can be useful to help identify the outcomes 8 HHH HHT HTH HTT THH THT TTH TTT is the sample space This is the listing of all possible outcomes of three tosses of a coin Determine the event Ethat at least two heads appear EHHHHHTHTHTHH Determine the event Fthat at exactly one head appears F HTT THT TTH Example 5 Two marbles are selected from a bag containing 4 red marbles 3 blue 039 O 0 marbles and 1 green marble The colors of the marbles and the sequence in which they were pulled are noted Describe an appropriate sample space S for this experiment Describe the event Ethat a red and a blue marble are chosen E Describe the event Fthat both marbles are the same color F Describe the event G that the first marble is blue Consider What would a tree diagram look like for this experiment Example 6 Two marbles are selected from a bag containing 4 red marbles 3 blue marbles and 1 green marble The colors not order are noted in C39 O O 390 Example 7 a Describe an appropriate sample space S for this experiment Consider Can a tree diagram be used effectively Compare with 5 Describe an appropriate sample space S for this experiment Describe the event Ethat a red and a blue marble are chosen E Describe the event Fthat both marbles are the same color F Note the event G described in example 5 is inappropriate here An experiment consists of casting a pair of distinct dice perhaps one blue and one red and observing the numberthat falls uppermost on each die b Determine the events E2E3E6E7E11 that the sum ofthe numbers falling 0 O D uppermost is 2 3 6 7 and11 respectively Die 2 E21x1 I o o 0 o 0 E3 1221 0 I o o aquot E6 11 12 13 14 15 16 E7 1625lt34435261 E11 5 6 6 5 21 22 23 24 25 26 Determine the event F that both dice a display odd numbers and the sum is E 31 32 343 34 345 346 less than 7 F 41 42 43 44 45 46 Determine the event G that the product of 541 542 53 544 545 56 the numbers uppermost on the dice is 12 G 61 62 63 64 65 66 Note that a tree diagram could be used to illustrate this experiment but the chart shown above is much better Math 166 Lecture Notes F1IF2 Section F1F2 Simple and Compound Interest Simple Interest The formulas involving simple interest are 1 l Pn and 2 AP1n or APPn where l is the interest P is the principal ris the interest rate and tis the time in years and A is the accumulated amount These formulas are very simple so any unknown A P r or t can be solved for nicely with a little algebra and any calculator or by hand Compound Interest Earned interest that is periodically added to the principal and thereafter itself earns interest at the same rate is called compound interest With compound interest money grows fasterthan with simple interest Also for a given interest rate the more compounding periods the faster it grows n The formula for compound interest isA P 11 where A is the accumulated amount 71 P is the principal ris the annual interest rate n is the number of compounding periods per year and tis the time in years To solve directly forthe principal the formula is in P A 1 J One might also be interested in solving forthe rate ror the time t The n algebra involved with solving forA or P involves raising numbers to rather large powers The algebra involved with solving for ror t involves taking n I7 roots or working with logarithms Because of the tedious algebra involved at one time finance tables were used to help with these calculations We will be using the TVM solver function in our calculators and therefore we will bypass most of the algebra even without the use of finance tables Continuous compounding of interest Earlier it was mentioned that for a given interest rate money grows faster the more times per year it is compounded That means that compounding annually once per year is the wors compounding of interest Compounding semiannually twice per year is better Compounding quarterly fourtimes per year is even better Compounding monthly 12 times per year is even better yet Compounding daily 365 times per year might seem like the absolute best Imagine however compounding per hour 8760 times per year or per minute 525600 times per year or per second 31536000 times per year Now imagine compounding infinitely many times per year and that is the concept of continuous compounding The formula isA 2 Pen where A is the accumulated amount P is the principal ris the annual interest rate compounded continuously and tis the time in years The formula for continuous compounding is straight forward enough to be able to solve for A and P quite easily algebraically without a special calculator function However solving for r or trequires working with logarithms The Excel worksheets below illustrate how compound interest works yean Note inte39rest per period is Cumulative Period Begin Balance Interest Ending Balance Interest Earned 1 1 1000000 30000 1030000 30000 1 2 00 309 00 Compound Interest Illustrator 1 year compounded quarterly Present Value investment 1000000 Number of Years 1 Note interest per period is llnterest Rate in Percent 600 600 4 Number of compounding periods 4 Cumulative Year Quarter Begin Balance Interest Ending Balance Interest Earned 1 1 1000000 15000 1015000 15000 1 2 1015000 15225 1030225 30225 1 3 1030225 15453 1045678 45678 1 4 1045678 15685 1061364 SB 61364 Compound Interest Illustrator 1 year compounded monthly Present Value investment 1000000 Number of Years 1 Note interest pe r39period39is Interest Rate in Percent 600 60012 Number of compounding periods 12 Cumulative Year Month Begin Balance Interest Ending Balance Interest Earned 1 1 1000000 5000 1005000 5000 1 2 1005000 5025 1010025 10025 1 3 1010025 5050 1015075 15075 1 4 1015075 5075 1020151 20151 1 5 1020151 5101 1025251 25251 1 6 1025251 5126 1030378 30378 1 7 1030378 5152 1035529 35529 1 8 1035529 5178 1040707 40707 1 9 1040707 5204 1045911 45911 1 10 1045911 5230 1051140 51140 1 11 1051140 5256 1056396 56396 1 12 1056396 5282 1061678 61678 Use the TVM solver when appropriate to solve the problems below Example 1 Find the accumulated amount future value AND the interest earned after 10 years if 20000 is invested at 45 nominal stated annual rate a at simple interest A P 1 rt A Interest b compounded annually N A I FV PV P Y Interest PMT C Y c compounded semiannually N A I FV PV P Y Interest PMT C Y d compounded quarterly N A I FV PV P Y Interest PMT C Y e compounded monthly N A I FV PV P Y Interest PMT C Y f compounded daily N A I FV PV P Y Interest PMT C Y g compounded continuously A Pert A Interest Example 2 Find the present value the amount that needs to be invested now in order to accumulate to 30000 at the end of 10 years in an account that earns 6 interest a at simple interest A P 1 rt p b compounded monthly N FV PV P Y P PMT C Y c compounded daily N FV PV P Y P PMT C Y d compounded continuously A Pert p Example 3 You deposit 3500 in a savings account paying 44 per year compounded semiannually After how many years will you have 5900 in the account N FV PV PY PMT CY years Example 4 Scott has narrowed his investment options down to two choices Examine them and determine which option will optimize his investment 1 Invest in an account that earns 765 compounded daily N PV PMT FV P Y C Y 2 Invest in an account that earns 77 compounded quarterly N PV PMT FV PY CY Effective Rate of Interest In 1968 Truth in Lending Act passed by Congress requires that the effective rate of interest be disclosed in all contracts involving interest charges giving consumers a common basis for comparing the various rates quoted by different financial institutions 71 The formula is re 1 1 1 where re t is the effective rate of interest ris the 71 annual nominal interest rate and n is the number of compounding periods per year Once again our calculators provide us with a convenient function to use Example 5 Find the effective rate of interest to 2 decimal places corresponding to a nominal rate of65 per year compounded a annually b semiannually 0 monthly d daily m You could now use the Effective Rate formulafunction to rework Example 4 Example 6 Melissa invested 3000 in an IRA and no deposits or withdrawals were made forthe next 4 years The interest was compounded monthly At the end of the 4 year period Melissa had 372150 in the account Find the effective rate at which Melissa s account was earning interest over that period How can you verify that your answer is correct Step 1 Find the nominal rate N l PV PMT FV P Y C Y Step 2 Find the effective rate Step 3 Verify the effective rate N PV PMT FV P Y C Y Math 166 Lecture Notes L1 Section L1 Introduction to Logic A proposition or statement is a declarative sentence that is either true or false but not both Note that commands questions exclamations opinions and ambiguous sentences are not statements Example 1 Which ofthe following are propositions statements a Watch out there s a snake Is is not b Is green your favorite color is is not 0 Children don t like broccoli is is not d Broccoli is considered a vegetable that is good for you is is not e Air and water are necessary for human life is is not f Dodo birds are extinct is is not 9 Dinosaurs roamed the earth at the same time as man is is not h 6X3 12 is is not I Mrs Lackey like asparagus is is not In logic conclusions must be inescapable Therefore every concept must be clearly defined so that there are no ambiguities and no vagueness In our problem sets we will limit our study to declarative sentences that are generally recognized to be either clearly true or clearly false Connectives Statements that express a single thought are called simple statements Simple statements combined with connectives such as and and or create compound statements Order of operations for connectives As with other mathematical operations there is a speci ed order ofoperations for connectives That order is OI39 v or i a on The symbol v represents an inclusive disjunction The symbol 1 represents an exclusive disjunction In order to do further work with logical statements and connectives we must be able to translate back and forth between written or verbal statements and the associated mathematical symbols We will not be studying these in depth in our course Example 2 Let p Sarah is five years old Let q Sarah will go to kindergarten this year Translate the following from symbols into words a p Sarah is ve years old b pAq Sarah is ve years old she will go to kindergarten this year c p v q Sarah is five years old she will go to kindergarten this year d plq Sarah is five years old Sarah will go to kindergarten this year e p a q Sarah is five years old she will go to kindergarten this year Example 3 Let p Sarah is five years old Let q Sarah will go to kindergarten this year Translate the following from words into symbols a Sarah is not five years old or she will go to kindergarten this year b Sarah will go to kindergarten this year and she is five years old 0 Sarah will either be five years old or she will go to kindergarten this year d It is not true that Sarah is five years old or she will not go to kindergarten this year e Sarah is neither five years old nor will go to kindergarten this year Note In meaning this is equivalent Sarah is not five years old and she will not go to kindergarten this year f If Sarah will go to kindergarten this year then she is five years old Truth or Falseness of Statements If we know the truth or falseness of statements p and q then we can also determine the truth or falseness of other logical statements involving these two propositions Example 4 Considerthe propositions defined in Example 2 p Sarah is five years old q Sarah will go to kindergarten this year Suppose that we know that Sarah is indeed five years old and she is going to kindergarten this year In other words statement p is true and statement q is true We can now examine the truth or falseness of the logical statements below a p Sarah is not five years old lsthisstatement T or F b pq Sarah is five and she is not going to kindergarten this year lsthisstatement T or F c p v q Sarah is not five or Sarah is not going to kindergarten this year lsthisstatement T or F d p l q Either Sarah is not five or she is going to kindergarten this year Is this statement T or F Example 5 Let p be the statement Daisies are a kind of ower Let q be the statement Green is a primary color Evaluate whether the given statements below are true or false 3 l0 q b I0 v q Truth or Falseness of Statements continued If we don t know the truth or falseness of statements p and q then we have to examine all possibilities Math 166 Lecture Notes 15 Section 15 Rules of Probability Odds vs Probability Properties of the Probability Function and Their Applications Property 1 PE 2 0 for any event E and PE S 1 for any event E Property 2 PS 1 where S is the sample space Property 3 Union Rule le and Fare any two events ofan experiment then PEUF PEPF PEmF Property 4 If E and F are mutually exclusive then PE UF PE PF Mutually exclusive means only one of them can occur also that E n F Q Property 5 Rule of Complements le is an event and Ecis the complement of E then PEC1 PE and PE1 PEC DeMorgan s Laws PEC UFO PEnF and PEC NFC PEUF Example 1 Applying PE U F PE PF Note We actually applied this concept in Example 10 of 14 Notes The superintendent of a school district has estimated the probabilities associated with the SAT verbal scores of students from that district The results are shown below Score X Probability gt 600 lt X s 700 500 lt X s 600 lt lt XS300 A student is selected at random What is the probability that his SAT verbal score will be a More than 400 b Less than or equal to 500 0 Greater than 400 but less than or equal to 600 Example 2 Applying PE U F PE PF e PE n F A card is drawn from a wellshuffled deck of 52 playing cards What is the probability that it is a queen or a heart Note As illustrated in the above examples when you are asked to find the union of two events it is important to recognize whether or not the events are mutually exclusive so that you know whetherthere is an intersection to deal with Example 3 The qualitycontrol department of a pickup manufacturer has determined from records obtained from the company s service centers that 35 ofthe pickups sold experience transmission problems 2 experience radiator problems and 08 experience both transmission and radiator problems before the expiration of the warranty a Find the probability that a pickup purchased by a consumer will experience transmission or radiator problems before the warranty expires Hint Apply PE U F PE PF 7 PE n F b What is the probability that a pickup bought by a consumer will not experience transmission or radiator difficulties before the warranty expires Hint Apply PEC1 PE Computations Involving the Rules of Probability Example 4 Let E and F be two mutually exclusive events where PE 3 and PF 5 Compute a PE n F b PE u F Q PE3 d PEC u FC Example 5 Let E and F be two events ofan experiment with sample space 8 Suppose PE 5 PF 4 and PEnF 25 Compute a PEUF b PECnFC m PEUFC c PECnF Example 6 Let S v w x y 2 be the sample space for an experiment where Pv 1 Pw 2 Px 15 and Pz 25 Let E w x y and F y 2 Compute a PM b PE m F c PEC nFC Odds vs Probabilities Odds and probabilities are related however the terms are not interchangeable Observe the vocabulary and relationships in the examples below Example 7 What is the probability ofthrowing a fair 6sided die and getting a three What are the odds in favor ofthrowing a fair 6sided die and getting a three 1 to 5 Note you could also express the odds as a ratio fraction 15 though this may be more easily confused with probability What is the probability ofthrowing a fair 6sided die and NOT getting a three What are the odds against throwing a fair 6sided die and getting three 5 to 1 Changing Probability to Odds f PE is the probability of an event E occurring then 1 The odds in favor of the occurrence of an event are given by the ratio of the probability ofthe event occurring t2 the probability of the event not occurring The odds in favor of E occurring are PE PE PE 7 1 1 PE PEC The odds against the occurrence of an event are given by the reciprocal ofthe odds in favor ofthe occurrence of the event In other words the odds against the occurrence ofan event are given by the ratio of the probability ofthe event not occurring to the probability ofthe event occurring 1 PE PE PE PE Note Whenever possible odds are expressed as ratios of whole numbers fthe odds in favor of E are ab we say the odds in favor of E are a to b fthe odds against E occurring are ba we say the odds against E are b to a l The odds against E occurring are PE 7 0 Example 8 The probability ofa certain event E is a Find the odds in favor of the event b Find the odds against the event Example 9 PE is 024 a Find the odds in favor of E b Find the odds against E Changing Odds to Probability fthe odds in favor of event E occurring are a to b then the probability of E occurring is PE a ab Example10 a You nd a shoe style you like The odds of nding it in your size are 3 to 1 What is the probability you will find the shoe in your size b The odds that the Cougars will beat the Lions in the next game are 3 to 4 fties are not possible what is the probability that the Lions will beat the Cougars Math 166 Lecture Notes 24 Section 24 The Binomial Distribution This section of material brings us to a discussion of a certain class of experiment When this type of experiment is involved certain probabilities can be computed using special formulas and calculator functions Caution Not all experiments fall into this category But for those that do the calculations learned in this section are much simpler than the calculations learned in earlier sections of material Bernoulli Trials In general experiments with two outcomes are called Bernoulli trials A sequence of Bernoulli binomial trials is called a binomial experiment A binomial experiment has the following properties 1 The number of trials in the experiment is fixed 2 There are two outcomes ofthe experiment success and failure 3 The probability of success in each trial is the same 4 The trials are independent of each other Probabilities in Bernoulli Trials Let Xbe the variable that gives the number of successes in a binomial experiment In this case the variable X is called a binomial random variable and the probability distribution oins called a binomial distribution Let p probability of success in any trial and q the probability of failure Note q 1 p To find the probability of exactly X successes in n independent trials we compute PX x Cnxpquotqquot x 0 1 2 n Example 1 A fair fourcolored spinner red blue green yellow is spun five times Compute the probability of obtaining exactly two blues in the five spins Solution 1 Construct a tree diagram or list the events in the sample space too big Solution 2 Compute the probability using counting methods and PE nEnS CS 2333 1027 s 2637 44444 1024 Solution 3 Treat it as a binomial experiment using PX x Cnxpquotqquot Note The number oftrials is fixed ve spins There are two outcomes blue is success not blue is failure The probability of success on each spin is the same The trials are independent of each other the results of one spin doesn t affect the next We need to know the values of n x p and q forthis problem n 5 trials spins x 2 successes 2 blues p i 1 out of4 sections is blue q 3 So PX x Cnxpquotqquot C5 2 x 252x 75 3 10 x 0625 x 421875 e 2637 Solution 4 Again treating it as a binomial experiment we will learn how to use a calculator function that performs the formula s computations for us PX 2 binom pdf n p x binom pdf 5 25 2 m 2637 Example 2 A fair vesectioned spinner with numbers 1 through 5 is spun seven times Find the probability of getting an odd number at least five times Solution 1 Draw a tree diagram write out the entire sample space and count Solution 2 Compute the probability using counting methods and PE nEnS C7 53333322C763333332C7737 5555555 5 6 7 W 2 32805 E 419904 4199 57 78125 Are there other options Can we solve this as a binomial experiment Note The number oftrials is fixed seven spins There are two outcomes odd is success even is failure The probability of success on each spin is the same The spins are independent of each other We want the probability of obtaining exactly 5 or exactly 6 or exactly 7 odd numbers What is X What is n What is p What is x What is q n mathematical symbols we want PX5 or X6 or X7 or PX5 PX6 PX7 Solution 3 Remember PX x Cnxpquotqquot We would need to work this out for each value then add PX 5 C7 56542 21 x 07776 x 16 2612736 PX 6 C7 66641 7 x 046656 x 4 1306368 PX 7 C7 76740 1 x 0279936 x1 0279936 4199040 m 4199 Solution 4 There is also a calculator function that can give you the values above Use the 2nd function button and find the DISTR key option 0 binomgdf PX 5 binom pdf 7 6 5 2612736 PX 6 binom pdf7 6 6 1306368 PX 7 binom pdf7 6 7 0279936 4199040 m 4199 Solution 5 There is another calculator function that can help you in this type problem Use the 2nd function button and find the DISTR key option A binomgdf PX50r6or71 PXOor1or20r3or4 1 binom cdf 7 6 4 1 580096 419904 m 4199 Example 3 As a oartrcutar dommg rtem comes ott tne assemow Mne rt nas a 3 cnance of oerng detectwe tr a tot of 12 dozen of tnese rtems rs snrooed to Betty s Bouuque SohmorM You coutd draw a tree dragram out rtwoutd have 144 orancnest Sotutron 2 You coutd use courmng tecnnroues andyou snoutd be aote to do that Does rt regresent a ornomrat exgenment Tne numoer ottnats ts xed t2 dozen rtems T ere are two outcomes efectwe ornot detectwe Tne probabmty of success on eacn rtem rs tne same Tne detects rn two rtems are mdependent oteacn otner we wru assume Sotutron 3 Workme rootem as a ornomrat expenment usrng tnetormuta rtyou nad a srmote screntrnc catcu ator and not onewrtn a specrat tunctron or programs Sotutron 4 Work rt as a rnomrat expertment usrng btnomgdf amp severat steps someumes you nave tot Sotutron 5 Worktt as a ornomrat expertment usrng ornomcdt oertecttortnrs orootemt We want tne probabmty otnndrng exac y 0 or exacuy1 or exacuy 2 detectwe rtems WhattsX Whaltsn Whattsp wnatrsm Whattsq Brnomcdt rs oertecttortnrs prootem 2 tunctron button Dtsw key ootron A binomgdf PX0ortor2 PXSZ bmom gdttmzt 03 2 a 1904931599 a 1905 nu ExampleA Afatrmreercotored spmner red b ue green rs spun 3 trmes L x be tne number of reds tnat occur rn tne tnree sorns Construct a probabmty drstnoutron tor x Sotutron My order to construct a probabmty dtsmbuuon we need to know aH tne oossrote outcomes at vatues are oossrote or X Tnen we need to nndtne probabmty assocrated wrtn eacn oossrote outcome x P1X Example 5 Of every 100 clothes dryers built at a particular factory 78 of them will not have problems that require repair until after the 1 year manufacturer s warranty is past Cindy recently purchased 20 clothes dryers from that factory for her laundromat What is the probability that 1 year from now a None of Cindy s machines will have needed repair b Exactly five of Cindy s machines have needed repair 0 At least seventeen of them will not have needed repair d At least two of Cindy s machines will have needed repair Solution Is this a binomial distribution If so what is X What is n What is X What is p What is q We should note that it is often helpful especially in the more challenging problems to create a quick table correlating the s of failures and the s of successes Ibad I 0 I 1 I2 I 3 I 4 I 5 I 6 I 7 I 8 I 9 I10I11I12I13I14I15I16I17I18I19I20I Igood2019I18I17I16I15I14I13I12I11I10I 9 I 8 I 7 I 6 I 5 I 4 I 3 I 2 I 1 I 0 I a It is worth noting that each of the above problems could be worked with success and failure defined exactly the opposite of the ways worked above a Math 166 Lecture Notes G1 Section G1 Game Theory Decision Making Game theory is a relatively new branch of mathematics In 1994 the Nobel Prize in Economics was awarded to several men for their work in this field Game theory is not just about games It refers to situations of conflict between participants The participants can be individuals companies armies governments or nonhuman entities such as the weather It combines matrix mathematics with probability theory and determines optimal strategies to opponents in competitive situations Payoff matrices are used to represent the strategies and outcomes of the players One participant s strategies will be represented on the rows and the other s strategies will be represented on the columns It is important to realize that the numbers in the matrix cannot be positive and negative at the same time in order to represent the viewpoint of both players Therefore we will understand that the values in the matrix represent the winnings of the row player and the losses of the column player Note Not all payoff matrices are square The number of rows andor columns depends on the number of moves or choices available to each player Carl s moves Carl s moves 1 2 3 1 2 3 1 5 0 1 1 5 0 1 Rachel s moves 2 3 4 0 Rachel s moves2 3 4 0 3 2 6 2 3 2 6 2 Rachel s perspective Carl s perspective Let s examine the payoff matrix above If Rachel s move is she could lose 3 win 4 or win 0 depending on Carl s move If Carl s move is then he will win 1 0 or 2 depending on Rachel s move From Rachel s perspective the 6 is the largest entry in the matrix and that is in row 3 She could choose row 3 but in order for herto get 6 Carl will have to choose column 2 Is Carl likely to choose column 2 No column 2 is actually his worst column Knowing that should Rachel make a different choice From Carl s perspective column 3 looks the best no losses and only 1 under his highest value in the game of 3 lf Rachel realizes this and expects him to make that move then what move should she choose She should choose move to minimize her losses But what is Carl thinking He could be thinking Rachel thinks I am going to choose column 3 so she will choose row 2 Therefore I ll choose column 1 and beat her at her own game and win 3 But maybe Rachel figures he might think that way and she thinks He thinks he is going to trick me by choosing column 1 ratherthan column 3 Therefore I ll choose row 1 and trick him and win 5 Now what if Carl figures out her thinking and on and on It could get pretty complicated And in a real game situation who knows what choices might be made especially when one is trying to outthink the other Each player must determine his strategy for the game If a human is playing against a nonhuman entity he will make one of two types of decisions The optimist will make a decision based on the best of everything happening The pessimist will expect the worst to happen and make the best of it As we begin studying basic principles of game theory we will make certain assumptions about the way players will choose their moves Here is the principle that we will adopt In a twoperson game each player wants to find a strategy that will maximize his winnings while minimizing his losses We will assume that each player follows this optimum strategy Still it is possible that once a player knows the strategy that the other player is using heshe may want to adjust their strategy ifthere is a benefit to doing so Sometimes knowing the opponent s strategy is of no benefit to either player A game in which the neitherpayer can bene t from knowing his opponent s strategy is a strictlydetermined game In a strictlydetermined game there is a saddlepoint for the game and it represents the value ofthe game lfthe value is zero it is a m game Example 1 Consider the game illustrated by the matrix below Consider the choices of each player and the optimum strategies Carl s moves 1 2 3 1 5 0 1 Rachel s moves 2 3 4 0 this is from the row player s perspective 3 2 6 2 If Rachel moves first what is the best strategy for her in order to minimize losses Whatever row she chooses her opponent Carl will logically choose the worst reaction for her Rachel because that is the best for him If Rachel chooses move 1 Carl will choose move 3 1 value for Rachel lf Rachel chooses move 2 Carl will choose move 1 3 value for Rachel lf Rachel chooses move 3 Carl will choose move 3 2 value for Rachel Knowing this Rachel chooses move 1 to minimize her losses lf Carl moves rst what is the best strategy for him in order to minimize losses Whatever column he chooses his opponent Rachel will logically choose the worst reaction for him Carl because that is the best for her If Carl chooses move 1 Rachel will choose move 1 5 value for her If Carl chooses move 2 Rachel will choose move 3 6 value for her If Carl chooses move 3 Rachel will choose move 2 0 value for both Knowing this Carl chooses move 3 to minimize his losses lf Rachel and Carl play simultaneously they will each choose the optimal move knowing what the possibilities are for the other player Therefore Rachel will choose move 1 and Carl will choose move 3 Example 1I continued Is the game strictly determined If so nd the value ofthe game Carl s moves 1 5 0 1 Rachel s moves 2 3 4 0 this is from the row player s perspective 3 2 6 2 If Rachel chooses move 1 and Carl chooses move 3 Rachel will lose 1 Suppose Rachel knows that Carl is going to choose move 3 If she sticks to move 1 she will lose 1 If she switches to move 2 she will lose 0 She should change strategies Suppose Carl knows that Rachel is going to choose move 1 If he sticks to move 3 he will gain 1 If he switches to move 1 or 2 he will lose 5 or gain 0 He should not switch Since Rachel would benefit from knowing Carl s strategy the game is not strictly determined Marking a payoff matrix to determine strictly determined games and values To select the appropriate strategies while thinking actions through from each player s perspective we will 1 Select the row player s optimum move by identifying the smallest entry in each row circle them and then choosing the largest of those mark it with a triangle The corresponding row is the optimum best move for that player Put an asterisk on that move row number 2 Select the column player s optimum move by identifying the largest entry in each column underline them and then choosing the smallest of those mark it with an upside down triangle The corresponding column is the optimum best move for that player Put an asterisk on that move column number Using the above strategies each player is minimizing his opponent s wins while maximizing his own wins Example 1 againI using the marking instructions above Carl s moves 1 2 3 1 5 0 1 Rachel s moves 2 3 4 0 this is from the row player s perspective 3 2 6 2 fthe largest of the chosen entries for the row player is the same as the smallest of the chosen entries for the column player that entry is the saddlepoint a star is formed That does not happen in this game Example 2 Consider the game illustrated by the payoff matrix below Carl s moves 1 2 3 1 5 0 1 Rachel s moves 2 3 4 0 always from the row player s perspective 3 2 6 1 a Consider the choices of each player and the optimum strategies f Rachel moves first what is the best strategy for her in order to minimize losses Whatever row she chooses her opponent wi logically choose the worst reaction for Rachel because that is the best for him If Rachel chooses move 1 Carl will choose move 3 1 value for Rachel f Rachel chooses move 2 Carl will choose move 1 3 value for Rachel f Rachel chooses move 3 Carl will choose move 3 1 value for Rachel Knowing this Rachel chooses move 3 to maximize her situation If Carl moves rstand f Carl chooses move 1 Rachel will choose move 1 5 value for herself 5 loss for him If Carl chooses move 2 Rachel will choose move 3 6 value for herself 6 loss for him If Carl chooses move 3 Rachel will choose move 3 1 value for herself 1 loss for him Knowing this Carl chooses move 3 to minimize his losses f Rachel and Carl play simultaneously they will choose the optimal move knowing what the possibilities are for the other player So Rachel will choose move 3 and Carl will choose move 3 fthey do that Rachel gains 1 b Is the game strictly determined If so what is the value of the game Suppose Rachel knows that Carl is going to choose move 3 If she sticks to move 3 she will gain 1 If she switches to move 2 or 1 she will do worse She should not change strategies Suppose Carl knows that Rachel is going to choose move 3 If he sticks to move 3 he will lose 1 If he switches to move 1 or 2 he will lose more He should not change strategies Since neither would benefit from knowing the other s strategy the game is strictly determined The value ofthe game for the row player is 1 Example 2 againI using the marking instructions Carl s moves 1 2 3 1 5 0 1 Rachel s moves 2 3 4 0 always from the row player s perspective 3 2 6 1 fthe largest of the chosen entries for the row player is the same as the smallest of the chosen entries for the column player that entry is the saddlepoint a star is formed This game has a saddlepoint The value of the game forthe row player is 1 winnings WeekeineReview 2 11 12 13 143 11 Terms Symbols and Laws Set eiements roster notation setebuiider notation equaiitv or sets subset proper subset empty set union intersection disioint universai set compiement The union or sets A and 5 AU 5 is the set or aii eiements that beiong to either A or 5 or both txixe AorxeBorbom The intersection or sets A and 5 A n 5 is the set or aii eiements in common With the sets A and 5 AnBxxeAandxeB 39 i Two sets are disioint itthev have no eiements in common that is it Am B 25 it Uis a universai set and A is a subset of U then the set of aii eiements in Uthat are not in A is caiied the complement of A and is denoted A A s T if Uis a universai set and A is a subset of U then A A d AUA U e Amt 25 Commutative Laws Associative Laws Distributive Laws for sets De Morgan s Laws AUBC ns and was Us 11 Sample Problems L tetcutuniversaiTAireinsTand5tsaiveT a iiiustrate With a venn diagram b F Us U c FindAnES d Find 0 e FindAUEiC Using a venn diagram shade AC n BU CC 6 a 1 Let U denote the set ofall students currently enrolled at AampM and A X e U X is majoring in math B X e U X is minoring in physics C X e U X is majoring in chemistry Find an expression in terms ofA B and C for a The set of AampM students majoring in chemistry or minoring in physics but not majoring in math b The set of AampM students with a doublemajor in math and chemistry or minoring in physics Describe in words A U C0 d Describe in words A n B U C 0 Give an equivalent eXpression Give an equivalent eXpression 12 Terms Symbols and Laws lfA is a set then nA denotesthe number ofelements in the set For any two sets A and B nA u B nA nB 7nA n B there is also an extension to 3 sets 12 Sample Problems L Ih lf nA 6 nB 7 and nAuB 8then find nAnB Five hundred students were surveyed It was found that 420 liked pizza 430 liked hamburgers and 370 liked pizza and hamburgers a How many students liked pizza or hamburgers b How many students did not like either ofthese foods Draw appropriate Venn diagrams to identify the area representing a A BCnC b AnBC CC c AC BCnCC d AUBUC A survey of 150 library patrons was taken involving their reading ofthree types of books science fiction poetry and biographies The results are described below Use a Venn diagram to illustrate the problem and answer the questions that follow 27 read science fiction only 50 read poetry books 20 read science ction and poetry 92 read science ction or biographies 9 read poetry but not science fiction or biographies 47 read only one of the three types of books 5 read all three types of books How many of the patrons surveyed read a at least one ofthe three types of books b none of the three types of books 0 exactly 2 ofthe 3 types of books d science fiction 1314 Terms Symbols Laws Experiment outcomes sample space events simple events trial tree diagrams mutually exclusive probability certain event impossible event uniform sample space probability distribution theoretical probability empirical probability frequency relative frequency Let S be a uniform sample space Then Psl Ps2 Psn l for each of the simple events sls2s3sn in sample space S n v and for any event E simple or compound PE W of posSIble outcomes In S nS 1314a Sample Problems 1 Experiment Toss a 4sided die 1 2 3 4 and spin a 5color spinner R B G Observe the results Also Consider this an experiment a uniform sample space Describe the sample space associated with the experiment How many events are associated with this experiment How many simple events are associated with this experiment Give the event E an even prime number is observed on the die Give the event F the spinner does not land on a primary color Give a description ofthe event G R2 R4 Find E n F Find E o FC Find PB2 Find PE Find PF xt npr rhmapom IN A math class has 74 students registered for the course The room seats 100 The number of empty seats is recorded on a certain class day a What is an appropriate sample space for this experiment b Describe the event E that more than 10 students are absent 0 Describe the event Fthat 4 or fewer students are absent I An experiment consists of choosing from two wallets one containing 100 and one containing no money The amount of money in the wallet is observed removed and pocketed This is done three times each time any money removed from the wallet is replaced in succession and the sequence of results is observed Determine the sample space S ofthe experiment A tree diagram might be useful to help identify the outcomes b Determine the event Ethat exactly 200 is won total 0 Determine the event Fthat no money is won total d e m Determine the event G that 100 is won on the second selection Determine the event H that 100 is won on the second pick amp a total of 200 is won 5 An experiment consists of casting a pair of distinct dice perhaps one blue and one red and observing the numberthat fas uppermost on each die a Determine the event E that a two turns up on the blue die Also nd PE b Determine the event F that both dice turn up prime numbers Also find PF 0 Determine the event G that the product of the numbers on the dice is less than 18 Also find PG Math 166 Lecture Notes F4 Section F4 Amortizations and Outstanding Principals Amortization of Loans When one borrows money it is often paid in periodic payments over time so that at the end of the designated period the principle amount as been repaid as well as the interest As far as formulas go the basic formula works similar to an annuity The big difference is that in an annuity the interest is a bonus with a loan the interest is a penalty The formula for computing the periodic repayment R on a loan odeollars to be amortized over n periods with interest charged at the annual rate of r would be PX R quot 1 11 71 However with the TVM Solver the process is very much like what we were doing in the previous section Example 1 A sum of 30000 is to be repaid over a 4year period through equal installments made at the end of each year If an interest rate of 72 per year is charged on the unpaid balance and interest calculations are made at the end of each year determine the size of each installment so that the loan principal plus interest charges is amortized at the end of4 years N FV PV PY PMT CY Example 2 Verify the results from Example 1 by creating an amortization schedule 3000000 3000000 X 072 1 Example 3 The Smiths are planning to purchase a house for 150000 and need to borrow the full amount The bank is requiring a main loan and a piggy back loan The piggy back loan will be for 10 ofthe purchase price and will be for a 10year period at a 65 interest rate The main loan will be a 30year loan with a 6 interest rate The interest charges are computed monthly on the unpaid balance We assume the loans with interest must be amortized by the end of the terms How much will their loan payment be forthe next 10 years forthe remaining 20 years N FV PV P Y PMT CY N FV PV P Y PMT CY Payment for the rst 10 years the next 20 years Note House payments usually include an additional amount to cover property taxes and insurance Add an additional 300 per month to the previous figure and that would be a more realistic total for your payment due to the lending institution who then pay your taxes and insurance for you Example 4 The Carlsons have examined their budget and have decided that the most they can afford is a 1200 monthly house payment The credit union through which they plan to acquire their loan advertises an interest rate of 72 for either 20 or 30 years Usually a longer note carries a higher interest rate butfor comparison reasons we are using the same rate What is the maximum amount that the Carlsons can borrow with each option Option 1 20 years N PMT FV PV PY CY Compute the total paid for the house including interest Option 2 30 years N PMT FV PV PY CY Compute the total paid for the house including interest Example 5 Jack and Judy have been making monthly loan payments on their home for the last 5 years The purchase price was 140000 They made a down payment of 15 and took out a 20year conventional home mortgage on the unpaid balance The interest rate on their loan was 54 lfthe house is now appraised at 175000 how much equity do Jack and Judy have in their house Step 1 Compute the amount borrowed Step 2 Compute the monthly payment amounts N PMT FV PV P Y C Y Step 3 Compute the present value of the loan with 15 years to go N PMT FV PV P Y C Y Step 4 Compute the equity in the home now Sinking Funds A sinking fund is an account that is set up for a speci c purpose at some future date Example 6 Farmer Jones has decided to set up a sinking fund for the purpose of purchasing a cotton picker in 3 years He expects that the picker will cost 60000 lfthe fund earns 72 interest per year compounded quarterly determine the size of the quarterly installments the farmer should pay into the fund N PMT FV PV PY CY Example 7 Show the first year ofthe sinking fund schedule Math 166 Lecture Notes 44 Extra Section 44 Extra Identifying basic variables and free variables When you encounter a system of equations that has infinite solutions you need to select the appropriate number of parameters and assign them to the correct variables One way to do this is to identify the basic variables and the free variables in the problem 1 Circle the 1 s in each row that lead the nonzero elements Identify the columns where those circled 1 s appear and the variables those columns represent These are the basic variables 2 Identify the remaining variables These are the free variables You will need to assign parameters to these variables Let s look at some examples Example1 3x 2y6 1 2312 6x 4y 12 The rowreduced matrix is 0 0 I 0 I Identify the basic variables x Identify the free variables y Let t y Interpreting the rows back into equations we get X 23 y 2 Now replace y with t and solve for X getting X 2 23t The solutions are ofthe form 0 2 23 t Examplez x2y4z2 1 0 0H X y 22 1 The rowreduced matrix is 0 1 2 I I I Identify the basic variables x and y Identify the free variables z Let t z Interpreting the rows back into equations we get 1x 0 and y 22 1 Now replace 2 with t and solve for y getting y 1 2t The solutions are ofthe form 0 1 2t t 1 0 1 l 50000 Example3 x y z 100000 x y 32 0 The rr matrix is 0 1 2 5039000 12x 08y 042 10000 0 0 0 0 Identify the basic variables x and y Identify the free variables z Let t z Interpreting the rows back into equations we get x z 50000 and y 22 50000 Now replace 2 with t and solve for x amp y getting x 50000 z and y 50000 22 The solutions are ofthe form 50000 t 50000 2t t Example4 X 2y 3z4w 5 2X 4y 8214w 1 3z10w 8 A The resulting rowreduced matrix is OOl Ix OONlt Ol amN Identify the basic variables x and z and w Identify the free variables y Let t y Interpreting the rows back into equations we get X 2y 32 4w 5 and z 3w 2 andw2 Now replace w with 2 in the smaller equation giving z322 z62 z4 then substitute w2andz4andytinto X2y3z4w5 X2t34425 x2t9 The solutions are ofthe form X y z w 2t 9 t 4 2 Example5 X 2y z3vw 5 22 v 2w 8 v w 2 X y z v w 1 2 0 0 25 l 4 l The resulting rowreduced matrix is 0 0 1 0 15 l 3 0 0 0 1 1 3 Identify the basic variables x andzandv Identify the free variables y andw Lettyand S w Interpreting the rows back into equations we get X 2y 25w 4 and z 15w 3 and v w 2 Now replace y with t and replace w with s Solve for the remaining variables working backwards from the shorter equations to the longest giving v w2 v s2 v2s z15w3 z15s3 z3 15s X2y25w 4 X2t25s 4 x 4 2t 25s The solutions are ofthe form X y 2 v w 4 2t 25s t 3 15s 2 s s Math 166 Lecture Notes 22 Section 22 Combinations Combinations A combination is a subset of robjects taken from a set of n objects without any regard to the order in which the objects are selected The number of combinations of n distinct objects taken rat a time is given by l Cn r L where rs n rn r Combinations to compute nCr sometimes written Cn r 1 Enter n on your home screen 2 Press MATH cursor overto PRB and down to 3 nCr 3 Enter r 4 Press ENTER Example 1 Work these by hand and on your calculator C6 3 20 C10 2 45 C9 4 126 C4 4 1 m Combinations occur in problems regarding card hands committees and other circumstances where a rearrangement ofa particular group is still considered the same group By contrast permutations occur where a rearrangement within a particular group would change its nature into a different selection Examples For example a group of four people chosen to be President VicePresident Secretary and Treasurer constitutes a different leadership group with each reassignment oftitles within the particular subgroup permutation Four people sitting on a committee can play musical chairs all day long but it is still the same committee of four combination Example 2 A representative group of three from the nineman team will be chosen for a picture Find the number of different representative groups of 3 that can be chosen Solution The arrangement doesn t matter we are counting different groups This is a combination NOT a permutation From a group of9 we are choosing 3 The answer is C93 84 There are 84 different groups of 3 that could be created Example 3 A Senate investigations committee of four members is to be selected from a larger group often members Determine the number ofways this can be done Solution This calls for application ofthe combination principle A committee of four is the same committee no matter how many times they rearrange in their chairs The order they sit are called are listed etc is of no consequence From a group of 10 we are choosing 4 The answer is C104 210 There are 210 different committees of fourthat could be chosen from the 10 members Example 4 How many 5card hands can be dealt from a standard deck of 52 cards Solution With card hands like committee members you can rearrange them all day long and it is the same group the same card hand Rearrangements ofa particular group don t present a different option so the order is not significant We use combinations From a group of 52 we are choosing 5 The answer is C52 5 2598960 There are 2598960 possible 5card hands that could be dealt from a standard deck Example 5 Wesley goes shing and catches 5 cat sh 3 perch 4 bass and 7 flounder He decides to choose 2 catfish 1 perch 1 bass and 3 flounder for a sh fry for his friends In how many ways can those selections be made Solution This is a combination application with multiple tasks that call for the application ofthe multiplication principle There is also an assumption here that the fish are unique For example Each bass is different from the other 3 bass in size etc Task 1 Choose 2 catfish from the group of 5 C52 10 Task 2 Choose 1 perch from the group of 3 C31 3 Task 3 Choose 1 bass from the group of 4 C41 4 Task 4 Choose 3 ounder from the group of 7 C73 35 Total number of ways to make the selections 10 x 3 x 4 x 35 4200 Example 6 The members of a string quartet composed of two violinists a violist and a cellist are to be selected from a group of six violinists three violists and two cellists n how many ways can the string quartet be formed if one ofthe violinists is to be designated as the first violinist and the other is to be designated as the second violinist Solution This problem involves both permutations and combinations as well as the multiplication principle Task 1 Choose the 2 violinists order matters P62 30 Task 2 Choose a violist from the group of 3 C31 3 Task 3 Choose a cellist from the group of 2 C21 2 Total number of ways to make the selections 30 x 3 x 2 180 Example 7 A group of spellers are going to be competing in a national contest From the twentythree members of the Spunky Spellers club five members will be chosen to represent the club at the contest Two alternates will also be chosen In how many ways could the contestants and alternates be selected m a good argument can be given for the two alternates to be ordered However in keeping with the way the Tan book approaches alternates for juries we will not order the alternates Solution Task 1 Choose the 5 contestants C23 5 33649 Task 2 Choose the 2 alternates nowthere are only 18 spellers to choose from C18 2 153 Total number of ways to make the selections 33649 x 153 5148297 Example 8 the quorum concept The local math club has 18 members When official club business is conducted they must have a quorum of 12 members present How many different ways can the math club have a quorum of members Solution If 12 members constitute a quorum then in order to conduct business at least 12 members must be present That means that either 12 13 1415 16 17 or 18 members can be in attendance to have a quorum Computation 1 the of ways to have exactly 12 people present C18 12 18564 Computation 2 the of ways to have exactly 13 people present C18 13 8568 Computation 3 the of ways to have exactly 14 people present C18 14 3060 Computation 4 the of ways to have exactly 15 people present C18 15 816 Computation 5 the of ways to have exactly 16 people present C18 16 153 Computation 6 the of ways to have exactly 17 people present C18 17 18 Computation 7 the of ways to have exactly 18 people present C18 18 1 Note that these are NOT separate Tasks to be completed in succession as separate pieces in the completion ofone overall task The multiplication principle does NOT apply We ADD 18564 8568 3060 816 153 18 1 31180 possible quorums Note A problem stating that a majority of members need to be present would involve the same concept The difference is that you have to compute the number that makes a majority more than half Example 9 MampM Mortgage company has 12 new mortgages to process In how many ways can these new mortgages be distributed to four of the company s processors if each processor is to handle three mortgages Solution 12 mortgage packets must be placed on 4 desks 3 per desk Task 1 Choose the 3 mortgages to be put on processor 1 s desk C12 3 220 Task 2 Choose the 3 mortgages to be put on processor 2 s desk C9 3 84 Task 3 Choose the 3 mortgages to be put on processor 3 s desk C6 3 20 Task 4 Choose the 3 mortgages to be put on processor 4 s desk C3 3 1 Total number of unique ways to distribute the mortgages 220 x 84 x 20 369600 Cautions As you practice a variety of mixed problems that involve one or more of the multiplication principle permutations and combinations you might feel as if a dense fog has settled The only way out ofthe fog is to keep working problems until the air clears again You have to practice enough to lift the fog before the next exam A pointto remember Neitherthe permutation formula nor the combination formula handles repetitions fyou are allowing repetitions of letters digits etc you will be using the multiplication principle laying out blanks for example Example 9 may seem to involve repetition but it is a different problem entirely It is more related to the indistinguishable objects In fact that formula can be used in Ex 9 WeekrinrReviewrtQ 11 12 13 11 Terms Symbuls and Laws Set etemertts msterrtetattert setrbu der netatten Equahty nf sets subset prepersubset empty set unten trttersecttert dtstetrtt untversat set emptement m A A A A A A 5 ts the set nf aH etements that beteng ts the set nf aH etements te Etther A at B What tn nmmnn thh the sets A and B AuBXXEADrXEBDrthh AnBxxaAandxaB we sets ate disjoint fthey have rm etements tn nmmnn that ts t A n s z A A t t tn A ts aHEdthe complement um and ts de ated 1 1 xtxs mm A N U ts a untvetsat set and A ts a subset nf U than Ur e B At Cemmutattve Laws Asseetattve Laws Dtstrtbuttve Laws fur sets A d AvA ZU e AnA DEMnrgan sLaWs AvB A nB and AnB A vB Sample Pmblems L Letqkunversa Atetnsandstsatve U a tuusttate thh a Venn dtagram A b FmdAvB FtndAnB d Ftrtd B0 e F HdAkBc 5 Using a Venn diagram shade AC mBU CC 539 L Let U denote the set ofall students currently enrolled at AampM and A X e U X is majoring in math B X e U X is minoring in physics C X e U X is majoring in chemistry Find an expression in terms ofA B and C for a The set of AampM students majoring in chemistry or minoring in physics but not majoring in math b The set of AampM students with a doublemajor in math and chemistry or minoring in physics 0 Describe in words A U C0 Give an equivalent eXpression d Describe in words A n B U C Give an equivalent eXpression 12 Terms Symbols and Laws lfA is a set then nA denotesthe number ofelements in the set For any two setsA and B 704 U B nA nB 7704 n B there is also an extension to 3 sets Sample Problems If nA 6 nB 7 and nAuB 8then find nAnB Five hundred students were surveyed It was found that 420 liked pizza 430 liked hamburgers and 370 liked pizza and hamburgers IN a How many students liked pizza or hamburgers U b How many students did not like either ofthese foods L Identify the area ofthe Venn diagram representing aA BC C bA BC CC U U A v v A A cAC BC CC dAUBUC U U A V V v i A survey of 150 library patrons was taken involving their reading ofthree types of books science fiction poetry and biographies The results are described below Use a Venn diagram to illustrate the problem and answer the questions that follow 27 read science fiction only 50 read poetry books 20 read science ction and poetry 92 read science ction or biographies 9 read poetry but not science fiction or biographies 47 read only one of the three types of books 5 read all three types of books 9 U How many of the patrons surveyed read a at least one ofthe three types of books b none of the three types of books c exactly 2 ofthe 3 types of books d science ction 1314a Terms Symbols Laws Experiment outcomes sample space events simple events trial tree diagrams mutually exclusive probability certain event impossible event uniform sample space probability distribution theoretical probability empirical probability frequency relative frequency Let S be a uniform sample space Then Psl Ps2 Psn l for each of the simple events sls2s3sn in sample space S n v and for any event E simple or compound PE W of posSIble outcomes In S nS Sample Problems 1 Experiment Toss a 4sided die 1 2 3 4 and spin a 5color spinner R B G Observe the results Also Consider this an experiment a uniform sample space a Describe the sample space associated with the experiment b How many events are associated with this experiment 0 How many simple events are associated with this experiment 0 Give the event E an even prime number is observed on the die e Give the event F the spinner does not land on a primary color f Give a description ofthe event G R2 R4 9 Find EnF 339 Find E o FC Find PB2 j Find PE X Find PF A A math class has 74 students registered for the course The room seats 100 The number of empty seats is recorded on a certain class day a What is an appropriate sample space for this experiment Describe the event E that more than 10 students are absent E Describe the event Fthat 4 or fewer students are absent F C39 0 An experiment consists of choosing from two wallets one containing 100 and one containing no money The amount of money in the wallet is observed removed and pocketed This is done three times each time any money removed from the wallet is replaced in succession and the sequence of results is observed I a Determine the sample space S ofthe experiment A tree diagram might be useful to help identify the outcomes Determine the event Ethat exactly 200 is won total C39 0 Determine the event Fthat no money is won total d Determine the event G that 100 is won on the second selection G e Determine the event H that 100 is won on the second selection and a total of 200 is won H 5 An experiment consists of casting a pair of distinct dice perhaps one blue and one red and observing the numberthat falls uppermost on each die a Determine the event E that a two turns up on the blue die Also nd PE c Determine the event G that the product of the numbers on the dice is less than 18 Also find PG G PG Die 2 21 22 23 24 25 26 Die I 32 33 34 35 36 41 42 43 44 45 46 3 51 52 53 54 55 56

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