×

### Let's log you in.

or

Don't have a StudySoup account? Create one here!

×

or

15

0

15

# Class Note for MATH 3338 at UH

Marketplace > University of Houston > Class Note for MATH 3338 at UH

No professor available

These notes were just uploaded, and will be ready to view shortly.

Either way, we'll remind you when they're ready :)

Get a free preview of these Notes, just enter your email below.

×
Unlock Preview

COURSE
PROF.
No professor available
TYPE
Class Notes
PAGES
15
WORDS
KARMA
25 ?

## Popular in Department

This 15 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Houston taught by a professor in Fall. Since its upload, it has received 15 views.

×

## Reviews for Class Note for MATH 3338 at UH

×

×

### What is Karma?

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 02/06/15
Math 3338 Probability Fall 2006 Jiwen He Section Number 10853 http math uh eduj iwenhemath3338fa1106 html Jiwen He UrWErsity er Huustun jiwenhemam uh Edu Mam 333E Prubabmty FaH zane August 237 September 1 was Probability p115 23 Counting Techniques I Ordered Samples m mwen He u nnnn any at Huustun jwwenhemam uh Edu Mam 333E Prubabmty FaH zane August 237 September 1 was Probability p215 Elements of combinatorial analysis 0 Proposition The product rule of ordered pairs Given n1 elements 11 am and n2 elements b1 bnz there are precisely 711 X 712 distinct ordered pairs 1 bj containing one element of each kind Proof Arrange the pairs in a rectangular array in the form of a multiplication table with 711 rows and 712 columns so that 1 bj stands at the interection of the 2th row and jth column Then each pair appears once and only once and the assertion becomes obvious 0 Proposition General product rule of ordered multiplets Given n1 elements 11 anl n2 elements b1 bnz etc up to nT elements 001 new there are precisely 711 X 712 X r r r X nT distinct ordered rtuples onl biz wir containing one element of each kind Proof by induction If r 2 the assertion reduces to the product rule of ordered pairs Suppose it holds for r 7 1 so that in particular there are precisely 712 r r r nT r 7 1tuples bi2 wir containing one element of each kind Then regarding the r 7 1tuples as elements of a new kind we note that each rtuple ailL b wir can be regarded as made i2 y up of a r 7 1tuple bi2 wir and an element ail Hence by the product rule of ordered pairs there are precisely 711012 r r r na 71an r r r nT rtuples containing one element of each kind m JiWEn He University at Heustenuwennemam uh edu Mam 3333 Prunammy Fan mus August 237 September 1 was Probability p315 Examples 0 217 Given n1 12 plumbers P1 P12 and n2 9 electricians Q1 Q9 there are N 711712 12 9 108 possible way of choosing the two types of contractors Pi Qj o 218 There are 711 4 obstetricians Q1 Q4 and for each Qi there are 712 3 choices of pediatricians Pj for which 0 and Pj are associated with the same clinic Applying the product rule gives N 711712 43 12 possible choices 0 219 217 continued Given n3 5 appliance dealers There are N 711712713 1295 540 way to choose the three types of contractors P7 Qj Dk o 220 218 continued If each clinic has both 713 3 specialists in internal medicine and n4 2 general surgeons applying the product rule gives N 711712713714 4 332 72 possible ways to select one doctor of each type such that all doctors practice at the same clinic m uwen He UrWErSlty er Heustenuwennemam uh euu Mam 3333 Prubabmty Fan mus August 237 September 1 was Probability p415 Ideal Experiment 0 Important facts 0 The sample space together with probability distribution de ned on it de nes the ideal experiment 0 The nature of the sample points is irrelevant for our theory 0 The same sample space can admit of a great variety of different interpretation A number of situations in which the intuitive background varies all are however abstractly equivalent to the same sample space in the sense that the outcome differ only in their verbal description 0 Combinatorial product rule Let A1alyyan1y B2blyybn2yquot39 er001yy00nr A1x32xwxxlta1b2mgti1mi1r 71 Wehavethen A1 gtltB2 gtlt gtltXT A1 gtlt B2 gtltrw gtlt XT m Jiwen He UrWErslty er Heustunuwennemam uh euu Mam 3333 Prubabmty Fan mus August 237 September 1 was Probability p515 Ordered Samples 0 Sampling with replacement and with ordering Suppose that we choose r objects in succession from a set or population of n distinct objects a1 an in such a way that after choosing each object and recording the choice we return the object to the population before making the next choice This gives an ordered sample afsize r onl 042 air Setting n1 n2 r r r nT n in the product rule we nd that there are preciser N nT distinct ordered samples of the size r 0 Sampling without replacement and with ordering Suppose that we choose r objects in succession from a set or population of n distinct objects a1 an in such a way that an object once chosen is removed from the population This gives a permutation of size r which is again an arderedsample afsize r onl 042 air but now 01 aL are distinct and there are n 7 1 objects left after the rst choice 71 7 2 objects left after the second choice and so on Cleary this corresponds to setting n1 n 712 n 7 1 nT n 7 r 1 in the product rule Hence instead of 71 distinct samples as in the case of sampling with replacement there are now only Nnn71rwn7rl 2PTn distinctpermulalians of size r Denote P71 by n nn 7 1 r r r 1 We have 711 PM 7 H n 7 T JlWEn He UnlvErslty er Huustunjlwennematn uh edu Mam 3333 Prubability Fall mus August 237 September 1 mus Probability p615 Distribution of r balls in n cells 0 Suppose we place r distinguishable balls into 71 different cells Placing balls into cells amounts to choosing one cell for each ball Numbering both the balls and the cells let il be the number of the cell into which the rst ball is placed i2 be the number of the cell into which the second ball is placed and so on Then the arrangement of the balls in the cells is described by an ordered rtuple i1 i2 2 This is equivalent to sampling with ordering 0 Cell allowed to contain more than one ball With r balls we have T independent choices This is equivalent to sampling with replacement and with ordering Therefore r balls can be placed into 71 cells in 717 different ways 0 No cell allowed to contain more than one ball r S 71 Clearly there are 711 n empty cells originally n2 n 7 1 emply cells after one cell has been occupied and so on Then the arrangement of the balls in the cells is described by a permutations of size r Hence the total number of distinct arrangements of the balls in the cells is PK 0 A great variety of conceptual experiments are abstractly equivalent to that of placing balls into cells m JlWEn He UnlvErslty er Huustunuwennemam uh euu Math 3333 Prubablllty Fall mus August 237 September 1 mus Probability p715 Distribution of r balls in n cells cont 1 IIll Birthdays The possible con gurations of the birthdays of r people correspond to the different arrangements of r balls in n 365 cells assuming the year to have 365 days Dice The possible outcomes of the experiment of throwing a die r times or throwing r dice correspond to placing r balls into n 6 cells Coin When tossing a coin we are in effect dealing with only 71 2 cells Sampling Let a group of r people be classi ed according to ageor profession The categories play the role of our cells the people that of balls Ramdom digits The possible orderings of a sequence of r digits correspond to the distribution of r balls places into ten cells called 0 1 9 Accidents Classifying r accidents according to the weekdays when they occurred is equivalent to placing r balls into n 7 cells Elevator An elevator starts with r passengers and stops at 71 oors The different arrangements of discharging the passengers are replicas of the different distribution of r balls in 71 cells Sex distribution The sex distribution of r people Here we have n 2 cells and r balls JlWEn He UnlvErslty er Huustuniiwennemam uh Edu Math 3333 Prubablllty Fall mus August 237 September 1 mus Probability p815 IIll Exam ple Birthdays 0 Question There are r people gathered in a room What is the probability that two at least have the same birthday 2 What is the probability that at least one has the same birthday as you 0 Answer Denote the sample space by Q i2 24 where for k 1 r ik E 1 n 365 is the birthday of the kth person Then 365T Denote the complementary event AC all birthdays are different which is equivalent to sampling without replacement and with ordering The product rule gives AC PT365 So the probability that all birthdays are distinct is PAC and that two or more people have the same birthday is PA17 PAC17 1332335 For T 21PM 04927 for r 23 PA m 05243 Denote the complementary event AC The birthdays of others are different than my birthday which is equivalent to sampling with replacement and with ordering The product rule gives AC 365 7 1V So the probability that at least one has the same birthday as me is PA 17 PAC17 lfwe wantPAm12 we have T m 711092864365 w 25261 men He University at Heustenilwennemam uh Edu Math 3333 Prubablllty Fall zuue August 2amp7 September 1 zuue Probability p915 IIll More Examples 0 Generality We consider random samples of size r with replacement taken from a population of the n elements 01 an We are interested in the event A that in such a sample 01 aL no element appears twice that is that our sample could have been obtained also by sampling without replacement The product rule shows that there are 71 different samples in all of which PT satisfy the stipulated condition Assuming that all arrangements have equal probability we conclude that the probability of no repetition in our sample is A PT nnilrwnirJrl pAgt M The following concrete interpretations of this formula will reveal surprising features l Throwing a die six times the probability that all faces 71 6 turn up is 7 6 m 001543 extremely improbable 2 Elevator An elevator starts with r 7 passengers and stops at n 10 oors The P 170 m 006048 quite probability that no two passengers leave at the same oor is improbable 3 Random sampling numbers Consider the number 6 271828 Every succession of ve digites represents a sample of size r 5 for a population consisting of the ten digits 0 1 9 The probability that ve consecutive random digits are all different is 1 03024 JlWEn He UnlvErslty er Heustenuwennemam uh euu Mam 333E Probability Fall mus August 287 September 1 mus Probability p1015 23 Counting Techniques ll Subpopulations m mwen He u nnnn any at Huustun jwwenhemam uh Edu Mam 333E Prubabmty FaH zane August 237 September 1 was Probability p1115 Ill Subpopulation sampling without ordering 0 Any set of r elements chosen from a population of n elements Without replacement and Without ordering is called a subpopulalian of size r of the original population The number of such subpopulations is given by 0 Theorem A population of n elements has precisely on 7 n T rln 7 r subpopulations of size r S 71 Proof If order mattered then the elements of each subpopulation could be arranged in r distinct ways Hence there are r times more ordered samples of r elements than subpopulations of size r But there are precisely nn 7 1 r r r n 7 r 1 such ordered samples and hence jsut nn71rwn7r1i n r 7 rln 7 r subpopulations of size r Remark An expression of the form 01 is called a binomial cae icient often denoted by 7 gt instead of Of The number 01 is sometimes called the number afcambinaiians afn things taken r at a time without regard of order Jiwen He UrWErslty er Huustunrwennemam uh edu Mam 3333 Prubabmty Fan mus August 237 September 1 was Probability p1215 Bridge and Poker 0 Example 222 Bridge A bridge hand consists of any 13 cards selected from a 52card deck without regard to order There are 0 m 635 billion different bridge hands Suppose a bridge hand is dealt from a wellshuf ed deck ie 13 cards are randomly selected from among 52 probabilities and 0 possible outcomes are equally likely Let A the hand consists entirely of spades and clubs with both suits represented B the hand consists of exactly two suits Since there are 13 cards in each suit the number of hands consisting entirely of spades andor clubs ie no red cards is 0 m 10 millions One of these 0 consists entirely of spades and one consists entirely of clubs so A Of 7 2 Since there are 024 6 combinations consisting of two suits of which spades and clubs is one such combination using the product rule gives B 024 X Then 00000164 PB W 00000983 m 0 0 10 000 0 Poker There exisits 032 m 25 millions hands at poker Let PM A the hand consists of five different faces values These face values can be chosen in 0amp3 ways and corresponds to each card we are free to choose one of the four suits Using the product rule gives A 45 X 0amp3 and 5 13 PA 4jcxcsr 05071 5 JlWEn He UnlvErslty er Heusteniiwennemam uh Edu Math 3333 Prubablllty Fall mus August 237 September 1 mus Probability p1315 IIll More examples 0 Placing r balls in 71 cells There are 71 possible arrangements Let k S n and A a specified cell contains exactly k balls The k balls can be chosen in 0 ways and the remaining r 7 k balls can be placed into the remaining 71 7 1 cells in n 7 IV k ways using the product rule gives A C X n 7 1V4 It follows that A 7 Cgtltn71l k 1 1Tk 1314 Cgtlt7gtlt177 nT nT nk n The is a special case of the socalled binomial distribution Example 223 A technician selects 6 of 25 printers of which 10 are laser printers and 15 are inkject printers there are N 035 ways of doing it Let r S 6 and T exactly r of the 6 selected are inkjet printers There are 035 ways of choosing the r inkjet printers and then 03 ways of choosing the 6 7 r laser printers using the product rule gives AT 0amp5 X 03 Then PA3gt ll 3083 and A3 A4 A5 PA3 UA4 U145 PA3 PA4 PA5 T T T 853 JlWEn HE UnlvErslty er Huustumuwenhe mam uh euu Mam 333E Probability Fall mus August 237 September 1 mus Probability p1415 23 Counting Techniques Partitions m JiWEn He UrWErsity er Huustun jiwennemam uh Edu Mam 333E Prunammy FaH zane August 237 September 1 was Probability p1515

×

×

### BOOM! Enjoy Your Free Notes!

×

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

Jim McGreen Ohio University

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

Allison Fischer University of Alabama

#### "I signed up to be an Elite Notetaker with 2 of my sorority sisters this semester. We just posted our notes weekly and were each making over \$600 per month. I LOVE StudySoup!"

Bentley McCaw University of Florida

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

Parker Thompson 500 Startups

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

Become an Elite Notetaker and start selling your notes online!
×

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com