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Probability With Apps ISYE 2027
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Author Richard Serfozo Basic Probability Problems May 20 2003 Springer Berlin Heidelberg New York HongKong London Milan Paris Tokyo 1 Elementary Concepts The subject of applied probability appears to be rather fragmented be cause problems involving randomness arise in many different contexts and they often require the use of ad hoc mathematical techniques The subject has more underlying structure however than meets the eye The set of notes before you describes this structure via a number of basic probability problems These notes are intended as a supplement to an introductory prob ability textbook that describes the methodology and notation The basic problems presented here should be viewed as a followon to elementary motivating examples A reader just learning probability should aim at mas tering these basic problems in the sense of being able to recognize them in various settings and solve them by carry out the required analysis and com putations A good way to study each problem is to create and solve one or more examples of the problem analogous to those belowi By creating new examples this way one will actually own a piece of the subject as well as understanding it 11 Probabilities of Events A description of events and their probabilities requires a framework for rep resenting events of interest in terms of a random experimenti Our textbook tells us how to describe such an experiment in terms of a set of outcomes 0 called a sample space and to represent events as subsets of 2 A basic problem in this regard is as follows Problem 1 Sample Spaces and Events as Sets Given a verbal de scription of a random experiment and certain events of interest represent the sample space as a set 0 and identify the events as subsets o here may be several natural choices for 0 select the simplest one that contains enough information to describe the events of interest Be able to describe a 2 1 Elementary Concepts complicated event that may consist of intersections unions or complements of other events Example 1 1 Parts exiting an assembly line are inspected and categorized as being good or bad defective In a batch of 10 parts an inspector is interested in the following events A The number of bad parts is 0 B The number of bad parts is S 1 C The number of good parts is 2 5 The random experiment consists of observing a batch of ten parts Since each of the preceding events can be described by the number of bad parts in the batch we can represent an elementary outcome as the number of bad parts in the batch Accordingly we represent the sample space as Q 01 10 Then the events above are represented respectively as A0 B01 001m41 Example 12 In the preceding context suppose the inspector is interested in the following events associated with the rst ve parts A The rst two parts are good and the last three are bad B Three good parts are produced before two bad ones are These events cannot be described as above by the number of bad parts but are describable by the sample space Q1112I511i00rl where 1 means a good part and 0 means a bad part Then A 11000 and B 11100 01110 10111 10110 11011 11010 11101 11100 11110 11111 Upon de ning a random experiment the next step is to assign proba bilities to the outcomes and use these to compute probabilities of events Problem 2 Elementary Probability Computations Suppose you are considering an experiment with a probability space 0 that is discrete it is nite or countably in nite A probability measure P for such a discrete space is de ned by specifying the probability Pw for each elementary out come w such that Ewen Pw 1 Methods for doing this are discussed below Then you should be able to compute the probability of any event A which is PA 2 130 weA Example 21 Consider the random experiment that describes the number of re alarms that a certain station responds to in a week Assume that it is known based on historical data via the frequency method discussed 11 Probabilities of Ehents 3 shortly that the probabilities of exactly 01 2 3 4 5 alarm responses are 10l15120125120l10 respectively These numbers sum to 1 Then the event A that the number of alarm responses is at least 3 is PA Z Pu 25 20 10 55 406A Similarly P less than 2 alarm responses 10 115 25 The preceding examples were for nite or countable sample spacesl Many problems however involve sample spaces that are uncountable For instance suppose that one is interested in the probability that a worker completes a task within a certain time say within 30 minutes where the time may be anywhere in the interval 0 60 and the time scale is in min utes Then the sample space of this experiment of recording task completion times is Q 060 This space has an uncountable number of points We refer to the task completion time as a continuous variable since it takes values in an interval Other examples of continuous variables are tempera tures weights volumes stresses pressures and locations in the plane or a space where one is interested their values over intervals or rectanglesl For continuous variables we use the following general de nition of a probability measure A probability measure P on a sample space is a function that assigns a number PA to each event A in 0 such that the following properties are satis ed 0 g PA g1 PQ 1 and for any collection of disjoint events A1 1 l l An PUZ1Ak 213040 161 This de nition is a generalization of the one above for countable state spaces Problem 3 Probabilities for Uncountable Outcomes For an exper iment with an uncountable sample space and a probability measure on it compute the probability of an event Example 31 A aw is detected in an underground cable of length 4000 feet The aw is equally likely to be located anywhere which means that the probability of the aw lying in any region A C 0 4000 is PA length of A4000 This rule obviously satis es the de nition of a probability Find the probabilities of the following events B the aw is located within 200 feet of the center of the cable C the aw is located within 100 feet of either end of the cable By the rule for de ning P in this experiment it follows that 4 1 Elementary Concepts PB Length of the interval 1800 22004000 4004000 ll PC Length of the region 0100 U 390040004000 2004000 05 The next problem concerns how one obtains probabilities for random experiments Problem 4 quot 39 of P1 quot39 Select r 7 39139quot for the events of an experiment by one of the following methods Frequency Method Repeat the experiment many times and set PA equal to the percent of the outcomes in which A occurred the relative fre quency of A Logical Method Use logic of the experiment to determine the probabili ties Subjective Method Use intuition or a best guess of an individual or group of persons to assign the probabilities Later we will study some standard probability distributions eg the binomial or normal distributions that occur frequently in nature and would be appropriate for any of the preceding met odsi For most of our work the probability measure or distribution will be speci ed and our focus will be on computations with themi For applications that you encounter in other courses or in practice you may have to select the appropriate probability measure The frequency and logical methods are used the most The fre quency method is feasible when there is historical data available or when one can perform the experiment many times at a reasonable cost i The subjective method is used as a last resort when the other two ap proaches are not practical The subjective method is sometimes made rig orous by using utility functions of ones preferences for certain quantities and deducing the probabilities from the functions this is not treated in standard coursesi Example 41 A company is interested in forecasting the quantity of a product that it will sell in a month The aim is to nd the probabilities of selling 01i i i 5 units of the product no more than 5 per month have ever been sold typical of some expensive products The company has historical data of the quantities sold in the last 24 months The demands during this period have been stable no seasonal variations and it is assumed that the demand during a month does not affect the demands in other monthsi Therefore we can consider the quantities sold x1x2nixg4 in the 24 months as 24 samples of the random experiment of the quantity sold per month of the productlli Let pk denote the probability of selling h units where h 0i i i 5 Then by the frequency method we assign or estimate these probabilities as pk number of xils that equal h24i 11 Probabilities of Ehents 5 This is the percentage of zils that equal kl Speci cally suppose the quan tities sold are 343045222411024544121134 Then p0 224 1127 101 5247 p2 5247 p3 18 p4 724 p5 112 These would be be probabilities the company would use for forecasting For instance the probability of selling at least 4 units in the next month would be P4P5 38 Here are some examples of the logical methodi Example 42 Suppose an experiment has n outcomesr If we say that each outcome is equally likely then it follows logically from the de nition of a probability measure that the probability of each outcome must be 1ni Because of this the probability of an event A is PA1AlWL where 1A1 denotes the number of outcomes elements or points in the set A and 9 is the total number of outcomes which we assumed was nr Example 43 Three companies are bidding on a contract The relative qualities of the companies are such that Company A is twice as good as Company B and Company B is three times as good as Company Or What is the probability of each company winning the contract This is an ex periment in which the elementary outcomes are A B C that the respective companies win the contract The logic of this setting says that PA 2PB PB 3PC where PA PB PC 1 Thus we have three equations in three unknowns and the solution is clear y PA 6 PB r3 PCr1r Example 44 A system fails when a defect occurs in one of 9 subsystems labeled 1i r r 9 only one defect occurs at a time Let pi denote the prob ability that the defect is in subsystem ii Suppose that each of the subsys tems 456 is twice as likely to contain the defect as any one of the other subsystems This information tells us that pi 210139 for i 6 456 and j g 4 56 for instance p4 2109 From these equalities it follows that 397 16 fori456 p17 112 otherwisei 6 1 Elementary Concepts The main property de ning a probability P for an experiment with outcome space 0 is as follows If an event A is the union of disjoint events A1An that is A A1 UAg UH UAn then PA 7 EMA i1 We also know that 139 l and 0 S PA S l for any event A C 2 These relations lead to further properties of P Here are a few HA 717 PA PAUB 7 PAPB7 PA B PA 7 PA n B 7 PA n B PA PB for A c B Let us see how these properties are used Problem 5 Algebra of Events and Probabilities If the probabilities of certain events are known how can one use these to compute the probabilities of other events Use the algebra of sets and basic properties of probabilities Example 51 A certain product was found to have two types of minor defects The probability that an item of the product has only a type 1 defect is 2 and the probability that it has only a type 1 defect is 2 Also the probability that it has both defects is 1 Find the probabilities of the following events A An item has either a type 1 defect or a type 2 defect B An item does not have either of the defects C An item has defect l but not defect 2 D An item has exactly one of the two defects Solution Let D1 D2 denote the events that the item has defects 1 2 respectively We know that PD1 2 PD2 3 and PD1 D2 1 Then by the de nitions of the events and the properties of probabilities it follows that PA 7 1311 u D2 7 PD1 PD2 7 1311 n D2 7 2 7 3 7 1 7 4 PB 7 1311 u DQV 717 1311 u D2 7174 7 6 PC 7 1311 n D3 7 PD1 7 1311 n D2 7 2 7 1 7 1 PD 7 PB 7 1311 n D2 7 4 7 1 7 3 The last expression uses the fact that B is the union of the disjoint events D and D1 D2 We now discuss random experiments with equally likely outcomes Al though these experiments have a simple probability structure they cover many diverse applications involving permutations and combinations of things The main difficulty is in deciding which counting principles to use to determine the numbers of outcomes in the events and sample space 11 Probabilities of Ehents 7 Problem 6 Probabilities Involving Equally Likely Outcomes If an experiment has equally likely outcomes how does one evaluate the probabil ity of an event In this case the probability of an event A is PM ViiWt where lAl denotes the number of outcomes elements or points in the set A We can evaluate these numbers by using the standard counting principles described below or by direct enumeration if a set is sma l The following are two fundamental counting principles which are also used in other engineering problems not involving randomness 61 Multiplication Principle Consider a job consisting of performing k tasks where the numbers of ways of performing the tasks are n1 i i i nk respectively Then the number of ways in which the job can be done is n1 nk This principle is used in counting ordered samples which are ktuples of the form 11 i i zkl Namely suppose there are n1 choices for 11 n2 choices for 12 Hi and nk choices for zki Then the number of ktuples 11 116 is the product n1nki Two important special cases of the multiplication principle are as fol lows Permutations The number of permutations of n items is nl nniln722li This is also the number of n dimensional vectors that can be formed without replacement from a population of size n Sampling Without Replacement The number of kdimensional vectors that can be formed without replacement from a population of size n is nn7 ln72n7 kl nlni kli These counting techniques have all been for counting numbers of vectors in sets We not switch to counting subsets instead of vectors 62 Combinatorial Principle The number of subsets of size k that can be formed from a set of size n is the binomial coe cient n 7 nl 7nniln7kll ltkgt imif This n choose kquot quantity is also called the number of combinations ofn things taken k at a time 8 1 Elementary Concepts This principle is used for counting unordered samples which are subsets of the form 1111 The number of such subsets that can be formed from n distinct objects is Keep in mind the difference between a ktuple and a set with k elements For instance 123 132 213 but 123 f 132 f 213 The following are four examples of combinatorial probability problems that use the counting principles Essentially all the combinatorial problems we will cover will be of these forms so study them carefully The setting of the examples will be a production facility that oper ates as follows Five types of parts labeled 1111 5 are produced for com panies X Y Z1 Each part is identi ed by its type and the company for which it is destined e1g1 part 3X is a type 3 part for company X1 A single production run consists of one part of each type for each company 1X1115X1Y111 5Y1Z 1 1 1 5Z1 The total number of parts by the mul tiplication principle is therefore 15 The following examples refer to this lot or population of 15 parts Example 61 Ordered Sampling with Replacement Suppose a random sample of four parts is chosen one at a time with replacement from the lot of 151 For instance the parts may be inspected and replaced at four stages in the production Find the probability of the events A The rst 2 parts are type 5 parts B The rst and last parts are of type 1 2 or 31 Solution These events involve the order of the parts and so de ne the sample space 0 as all 4tuples of parts chosen with replacement from 15 eg an outcome is 3Y5X1Z5X1 Since each sample is equally likely PA1Allf2 32152154 1251 Here 9 154 since 0 consists of 4tuples and there are 15 choices for each of the four entries Also lAl 32 152 since A consists of 4 tuples in which there are 3 choices for rst two entries and 15 choices for the last two entries Similar y PB lBlWl 92152154 9251 Here B consists of 4tuples in which there are 9 choices for the rst entry 15 choices for the second and third entries and 9 choices for the fourth entry1 Example 62 Ordered Sampling Without Replacement Suppose in the preceding setting that a random sample of four parts is chosen one at a time without replacement from the lot of 15 parts For instance an in spector analyzes four parts at once or a destructive test is done on four parts Find the probabilities of the events 11 Probabilities of Ehents 9 A All four parts are of type 12 or 3 B The rst and last parts are for company Zr Solution Since event B depends on the order of the parts we de ne the sample space 0 as all 4tuples of parts chosen from the lot of 15 without replacement Each sample is equally likely and so 987 6 PAlAll9l15141312 6651 Here there are 15141312 choices respectively for the four entries of an outcome in Q and there are 9 8 7 6 choices respectively for the four entries of an outcome in Ar Similarly 541312 7 PBlBll9l15141312 2211 Here there are 5 and 4 choices for rst and last entries of an outcome in B and there are 13 and 12 choices for the middle entriesi Example 63 Unordered Sampling 1n the preceding setting suppose four parts are randomly chosen without replacement from the lot of 15 Find the probabilities of the events A All four parts are of type 1 or 2 B Two parts are for X and two parts are for Yr C One part is for X one part is for Y and 2 parts are for Z Solution Since these events do not depend on the order in which the parts are chosen we de ne the sample space 0 as all subsets of four parts chosen from the lot of 15 Again each sample is equally likely and so PltAgt Amm 1gt 191 Here 9 is the number of subsets of size 4 that one can choose from the set of 15 parts and 1A1 is the number of subsets of size 4 that one can choose from the set of 6 parts of type 1 or 2 Similarly PB 13mm 20273 Here B consists of two subevents two parts are for X which has pos sibilities and two parts are for Y which has possibilities Then the number of outcomes in B by the multiplication principle is the product of these two binomial coef cients Using similar reasoning PC 10mm 5 5 50273 1O 1 Elementary Concepts Example 64 Permutation Problemi Suppose the lot of 15 parts described above are randomly put on a conveyor to go into a storage area Find the probabilities of the events A The rst three parts are for company Xi B The last four parts are of type 1 or 2 Solution De ne the sample space 0 as the set of all permutations of the 15 parts It therefore has 15 outcomes Since all outcomes are equally likely7 PA lAllm 54312x151 2915 Here lAl is the number of permutations of 15 parts With an X part in the rst 3 entries there are 5 4 3 choices for this and the number of choices for the remaining entries is 12L Similarly7 PB 13mm 6 54311x151 1915 2 Conditional Probabilities and Independent Events Many quot 39 involve A quotquot39 quot39 that some special event has occurred Assuming that an event occurs or will occur we have to evaluate probabilities of events differently than if we did not know the event occurred The new quotquotquot a quot39 1 quotquot 39 77 Problem 7 Conditional Probabilities From Initial Probability Mea sure Knowing the probability measure of an experiment how does one com pute the conditional probability of an event B given that A occurs Use the de nition PltBAgt M u BgtPltAgt provided PA gt 0 Example 71 An item is randomly drawn from a set of items abcd123IIIIIIVIVi Find the probability that item c is drawn given that it is known somehow that the item drawn is one of the rst 6 in the set Solution Let A denote the event that the item drawn is one of the rst 6 in the set and let B denote the event that item c is drawn Since each item is equally likely PA 12 Then PBlA PA BPA Pc is drawn from abcd l2PA 1612 13 lnstead of denoting events by A B etc we could have expressed the events in words For instance Pitem drawn is lllitem drawn is a roman numeral 7 PHI is drawn from 111111 VI V 7 Pitem drawn is a roman numeral 15512 1225 12 2 Conditional Probabilities and Independent Events Example 72 A customer promises to place an order sometime in the next 10 months for a product you produce You have to plan production for the next year and so you decide to come up with educated guesses maybe based on the customers hunches or past behavior for the probabilities p1p2i r r p10 that the customer places the order in months liHlO re spectively These probabilities sum to one since the order is assured to be placed The following are typical conditional probabilities that one can obtain from the preceding probabilities which are supposedly known To express events in a convenient manner we let X represent the month in which the order is placed The conditional probability that the order is placed in month 4 given that it is placed after month 3 is M 4X 2 4 PltX 4 n X 2 4PltX 2 4 104104 1010 The conditional probability that the order is placed after month 7 given that it is placed after month 2 is M 2 8lX 2 3 mm 2 s n X 2 3PltX 2 3 P3 109 P10P3 1010 The preceding example involves obtaining a conditional probability in terms of regular probabilities We will now discuss more interesting uses of conditional probabilities in which conditional probabilities are speci ed or they are easy to obtain and we want to nd regular probabilities of events or other conditional probabilities Problem 8 C quotquot P1 quot39 for Q j ovaents Com pute the probability of the intersection of the events A1 r r r An in terms of conditional probabilities PAilA1 Ai21 of the event Ai given the past events A1i r r Ai21 Use the multiplication formula PA1 n n An PA1PA2lA1 PAnlA1 n n AH Example 81 In a manufacturing plant there are 10 15 and 20 jobs of types I U and III respectively waiting to be processed on a machine eigi these may be physically different parts or orders for one type of product for different customers The machine operator randomly selects parts one at a time for processing ie each part is equally likely to be selected at any time What is the probability that the rst 4 jobs processed are of types U II III II and in this exact order Solution Let A denote the event of interest We can consider this event as A A1 Ag Ag A4 where A1 job 1 is type II A2 job 2 is type II A3 job 3 is type III and A4 job 4 is type If Then by the multiplicative property of conditional probabilities 2 Conditional Probabilities and Independent Ehents 13 PA PA1 A2 Ag A4 PA1PA2A1PA3A1 n A2PA4A1 n A2 n A3 1545144420431342 0072 Here 1545 is the probability of chosing one of the 15 type II jobs from the lot of 45 each job is equally likely Next PA2 A1 1444 since this is the probability of choosing one of the remaining 14 type 11 jobs from the lot ot 44 after one type II job was already chosen for job 1 The 2043 is the probability of choosing one job from the 20 type III jobs when there are 43 jobs left Finally 1342 is the conditional probability of choosing a type II job as job 4 given that there were already two type II jobs chosen from the 15 and so 13 remain Another approach to solving this problem is to treat it as a counting problem Namely there are 45 44 43 42 ways of selecting an ordered sample of four jobs from the 45 jobs and the number of ways that A can happen is 15 14 20 13 Thus 15 14 20 13 PM 45444342 quot072 Example 82 Consider a manufacturing system in which parts travel through a network of work stations as shown in Figure 21 The pjk7s are the conditional probabilities that a typical part traverses the arcs For in stance p35 is the conditional probability of going from station 3 to station 5 3 A 5 given that the traveler is at station 3 The possible departures from each station are assigned probabilities that sum to one eg p35p35 1 I724 1736 Fig 21 Traveling in a Network One is typically interested in the probability that a part takes a speci c route or that the part goes through a certain sector of the network or graph For instance using the multiplication rule for conditional probabil ities the probability of taking the route 1 A 3 A 5 A 4 is P1gt3gt5gt4P1H3P3gt5 1H3P5gt4 1gt3 3gt5 101310351054 l4 2 Conditional Probabilities and Independent Events Next consider the event A that a part goes through station 4 Clearly A is the union of three different routes that go through 4 Then reasoning as above we have PAP1H2H4P1H3H5H4P1H2gt3gt5gt4gt P121024 101310351054 1012102310351754 In the same way one can evaluate the probability that a part goes through a certain sector of the network by taking the sum of all the prob abilities of the routes that go through that sectorr As an exercise nd the probability that a part goes through 3 A 6 Also consider another such event and compute its probability Similar networks or graphs are also used to model a job or major task that consists of randomly evolving subtasksi Each node represents a subtask and the arcs leading out of a node lead to the next subtask that is chosen or determined by the probabilities on the arcsi lntuitively two events A and B are independen 77 if knowing that A occurs does not tell us anything about B and vice versa The mathematical de nition is PA B PAPBi We now describe three types of problems dealing with independent eventsi Problem 9 Independent Events Determine whether events A and B are independent or dependent Their independence can be established by verifying the de nition PA N B PAPB Another approach is to verify PAlB PA ii Establish the independence or dependence of a collection of events A1i r r An They are independent PAi1 quot39 Aik for any subsequence il l r r ik of l r r r n If any one of these equalities is not true then the events are dependent ii Be able to identify when to use these product formulas for independent events that may arise from physically independent phenomena Example 91 Suppose that a part is randomly selected from a set of parts 517M17L17527M27L2 which represents small medium and large parts 5 M L from two com panies l 2 Consider the events A the part is S or M 51M152M2 B the part is M or L M1L1M2L2 C the part is S 5152 D the part is from company 1 51M1L1 2 Conditional Probabilities and Independent Ehents 15 Describe whether or not these events are independent Solutioni Since each part is equally likely to be chosen PA46 PB46 PC26 13136 Now M n B PltM1 M2 26 y PltAgtPltBgt Thus A and B are dependent Next PC no P51 16 PCPDi Thus C and D are independent Note that all of the events A B C and D are not independent since we already know that A and B are dependent However are E and C independent How about A and C The next example illustrates the evaluation of a probability of intersec tion of independent events by multiplying probabilities of the events Example 92 A project consists of three independent tasks and the prob abilities of these tasks being completed on time are 90 80 and 95 respec tively Find the probabilities that i All three tasks will be completed on time ii The rst two tasks will be completed on time and the third one will not iii Either the rst or last task will be completed on time Solutioni Let Ci denote the event that task i is completed on time The physical independence of the tasks translates into the independence of the events C1 C2 Cgi Then by the de nition of independent events the rst two probabilities are 1301 n 02 n 03 9 s 95 0684 1301 n 02 n 0 capsules 0036 In the last equation we are using the property that independence of events also implies the independence of their complementary events The third probability of interest is P01 U C3 PC1 PC3 7 P01 C3 i9 95 7 i9i95 0995 Problem 10 TwoStage Experiments The outcome of an event A in volves a rst stage in which one of the events B1 i i i Bn leads to the event A at the second stage Compute the probability ofA by one of the following formulas 16 2 Conditional Probabilities and Independent Events n PA Z PA n Bk 161 HmiPWampWwM 161 The choice of formula would depend on what probabilities are speci ed or easily obtainable Example 101 The probabilities that a student in this class such as you will expend a high7 medium or low amount of effort in studying are 50 30 and 10 respectively Given that the student expends a high7 medium or low amount of effort7 the respective conditional probabilities of getting an A in this courses are 90 40 and 05 Find the probability that the student you will get an A in the course Let B17 B2 B3 denote the events of the student expending a high7 medium and low amount of effort Let A denote the event the student gets an A in the course Then 3 PA Z PABkPBk 90 X 50 40 X 30 05 gtlti10 i585 k1 This is too low for comfort lfl were you7 1 would study hard and then your probability of geting an A would be 90 Do it Problem 11 Bayes Formula Compute the conditional probability of an event by Bayes formula provided this approach is appropriate Namely events B17 i i i Bn form a partition of the sample space and you know the probabilities of these events and the conditional probabilities PAlBZ then Pwmmm meELPWMWBM One can usually identify when to use Bayes formula since the given information includes conditional probabilities like PAlBl and one is asked to nd a reverse conditional probability PBilAi An interpretation of Bayes is that A happens and we want to nd the conditional probability PBilA that Bi is the cause of Al Example 111 The members of a consulting rm rent cars from 3 rental companies 60 percent from company 17 30 percent from company 2 and 10 percent from company 3 The past statistics show that 9 percent of the cars from company 1 need a tuneup7 20 percent of the cars from company 2 need a tuneup7 and 6 percent of the cars from company 3 need a tuneupi If a rental car delivered to the rm needs a tuneup7 what is the probability that it came from company 2 Solution Let A denote the event that the car needs a tuneup7 and B17 2 Conditional Probabilities and Independent Ehents 17 B2 and B3 respectively denote the events that the car comes from rental companies 1 2 or 3 Then our assumptions are that PB1 0 6 PB2 0 3 PB3 0 1 PA B1 0 09 PA B2 0 2 PA B3 006 Therefore by Bayes7 formula we have 0 3oi2 0f PB2A 3 Random Variables and Their Distribution Functions Up to this point7 we have focused on stating events in words and comput ing their probabilities by considering these events as subsets of the sample space of all outcomes of an experiment This is a microlevel description of experiments7 Which is often not needed When focusing on measurable quantities Hereafter7 we Will consider events stated in terms of random quantities77 called random variables The probabilities associated With these random variables Will be represented by functions called distribution func tions h This is a higherlevel method of describing randomness that does not require the microlevel notation associated With a sample space This style of using random variables and distribution functions is standard for technical reports and articles involving probability and statistics 31 Discrete Random Variables Problem 12 De ning Random Variables Given an experiment and a random variable X of interest a realvalued quantity specify the random variable Xw as a function of the elementary outcomes w When X is a discrete random variable and the probabilities Pw are known or you can derive them obtain the probability function 101PXI 2 PW Manama This just involves summing the probabilities of all the elementary outcomes w for which the random variable equals the exact value I There are experiments in Which a random variable of interest is the same as the outcome of the experiment Here is an example Example 121 A trucking company is monitoring the amount of gasoline that each of its trucks uses A quantity of interest for a particular truck 20 3 Random Variables and Their Distribution Functions is the number of gallons X it uses on a particular 200 mile trip Then for the experiment that records the gasoline consumption on this trip in integer values a sample space of outcomes might be 0 1 21 1 1 50 This tacitly assumes that no more than 50 gallons is ever needed Note that for any outcome 4 the random variable Xw is the same as w eigi X42 42 In other words X is the identity function on 21 In such a case one usually denotes a generic outcome w by the letter 11 Then the probability function is simply px PXz Pz 1121H501 That is the probability function 101 of X is the same as the probability measure I 7 no extra work is needed to obtain The assumptions in this experiment did not mention anything about the probability measure P on the outcomes One way of getting this would be to use the frequency approach provided that one could repeat the trip about 20 times under essentially the same conditions Quite often a random variable is not the same as the outcome of an experiment and there may be more than one random variable associated with the experiment In these cases a little thought is needed to compute the probability function PX Example 122 A salesperson wants to assess future sales ofa slowmoving item in the next two months In the rst month the probabilities of selling 1 2 3 4 units of the item are 130 135 115 120 respectively In the second month the probabilities of selling 0 l 2 3 units are 125 130 140 105 respectively and the sales in one month have no bearing on the sales in the other month The salesperson is interested in the total number of sales in the two months Describe this as a random variable and specify its probability function Solution A typical outcome 8 of this random experiment is a pair such as 31 that represents the sales of 3 and 1 units in the respective months For this outcome the total number of sales in the two months is 4 and so X3 1 41 The probability of this outcome is P3l 115x30 01045 which follows by the independence of the two events of selling 3 and 1 units in the two months The values of Xw and Pw for all outcomes w are listed below Outcome w Xw Pw lOutcome w Xw Pw 1075 3 0 0375 10 1 3 11 2 109 31 4 1045 12 3 112 3 2 5 1060 13 4 1015 3 3 6 0075 20 2 0875 40 4 1050 21 3 1105 41 5 106 22 4 114 42 6 1080 23 5 0175 43 7 101 31 Discrete Random Variables 21 From this table we see that the possible values of X are 1r r r 7 Since we know the probabilities of the elementary outcomes we can compute the probability function of X by 1095 2 Pw 11m7 Manama This function is as follows I 6 7 171 i075 r1775 r2625 25 r1375 r0875 01 It is common to have several random variables of interest associated with an experiment For instance in the last example one might have been interested in the numbers X1 X2 of units sold in the two months or in the random variable Y that represents the minimum of X1 and X2 The probability functions of these can be obtained by the procedure we used for X do this as an exercise r The preceding example shows how to de ne random variables from in formation of an experiment This microlevel example is instructive for un derstanding random variables but for practical problems this level of detail is not needed The standard approach is to start at the macrolevel with probability functions or distributions and not bother with the underlying probability space We rst discuss discrete random variables and later dis cuss continuous random variables Problem 13 Probabilities of Discrete Random Variables Compute probabilities of interest for a discrete random variable X from its probability function 101 PX z or its distribution Fz PX S The basic ormulas are N e B 2pm 63 PaltX b Fb7Far Example 131 Suppose the number of emergency 911 telephone calls be tween 3 AM and 4 AM to a certain center is a discrete random variable X with probability function PX z e7337czl z 012Hr This is a Poisson probability function which we will study in depth later The probability of X taking on a single value is obtained by evaluating the preceding function at that value For instance PX 0 e73 n49 PX 3 3333 224 For more general probabilities one would like to use the distribution func tion 22 3 Random Variables and Their Distribution Functions 1 Fz PX S I 26733nnli 710 In this case this sum does not reduce to a simple formula and so we will compute probabilities using the probability function PX For instance 5 P2 lt X g 5 25533771 79 13 5 P2 g X g 5 25533771 977 12 Note that z 2 is not included in the rst sumi One may be interested in the probability that X is greater than some value If This is given by the in nite summation 5 PX gt z Z e SSnnli nx1 However it is easier to compute this via the complement formula a nite summation ac PX gt z 17 PX g z ZE SSnnli n0 For instance PX2117PX01ie 3i951 Also once some of these probabilities are known one can compute related ones eg knowing P2 lt X S 5 797 we have P2ltXorXgt517P2 X 5i203i In addition one can compute conditional probabilities just as we did for events For instance PltX 7lX 2 3 PltX 1X 2 3gtPltX 2 3 Plt3 X WM 2 s 7 2 Ze SSkkll 17 Ze SSkM i979 163 160 PX 2 7lX g 3 PX 2 7X g 3PX g 3 0 In many instances the probability function and probability distribution of a discrete random variable are not represented by formulas but are given by tablesi Here is an example of computing probabilities for a random variable from a table of its distribution function 32 Continuous Random Variables 23 Example 132 Consider the random variable X that represents the num ber of people who are hospitalized or die in a single headon collision on an interstate highway in a year The distribution of such random variables are typically obtained from historical data Without getting into the statistical aspects involved let us suppose that the distribution of X is as follows 0 1 2 3 4 5 6 7 8 Fzi250 i580 i784 i922 i972 i985 i991 i998 1000 Probabilities of X can be obtained by simply reading them from this table For instance PX 0 F0 25 PXgt217F2 17 784 i216 P3SXS6P2ltXS6 F67F2 i9917i784 207 Note that the last probability requires a little more care since we are dealing with the cumulative distribution A typical conditional probability is PX g M gt 2 P2 lt X g 5PX gt 2 W i452 lfthe probabilities for all values of X were of interest one would evaluate the probability function 101 PXz Fz7Fzil 11Hi8i A table of this function is as follows I l 0 1 pz i250 i330 i204 i138 i050 i013 i006 i007 i002 32 Continuous Random Variables In many instances one is interested in random quantities such as time distance pressure volume that take values in some interval of the real numbers Such continuous quantities are the focus of this section A random variable X is continuous if there is a nonnegative function fz called the density of X such that b Pa S X S b dz for any a S 12 31 The density f must necessarily satisfy 24 3 Random Variables and Their Distribution Functions fumzi Note that the distribution Fz PX S z of X can be written as Fz dy for any z E 700 312 From equations 31 and 32 we have PaSXSbFb7Fai 33 Taking the derivative of both sides of 32 we see that F 1 161 An important property ofa continuous random variable X is that the prob ability that it equals a specific value z is 0 That is PXxPzXz fydy0l Consequently PaltXSbPaSXSbPaltXltbPaSXltbl Problem 14 Probabilities of Continuous Random Variables Com pute probabilities of interest for a continuous random variable X from its density function by formula 31 or from its distribution function Fz by formula If is known compute F by If Fz is known compute f by Fz Example 141 Suppose X is a continuous random variable with the den sity function mw 73zamp and 0 elsewhere Find the following a The value of c b PX S 5 c PX S 3 d P4 S X S 55 e P4 lt X S 7 f PX S SlX S 35 Solution a To be a density function f should satisfy ff dz ll In this case 5 5 l c4371 dz ceg e71 dz cl 7 e73 3 3 Therefore c 17 34 10524 The probabilities in b7f can be determined by the probability distri bution of X and so we will determine it rst Clearly Fz 0 for z lt 3 Fzlforzgt6and 33 Percentiles of Distributions 25 ac Fz110524 AH dy 71 052417 43 3 g I g 61 3 Then the probabilities in b7f are as follows PX 5 7 F5 7 105240 7e 7 91 PX g 3 7 F3 7 0 P4 g X g 55 7 F5l5 7 F4 7 110524e71 7 25 7 301 P4 lt X g 7 7 F7 7 F4 7 17 F4 7 335 PX g 51X gt 4 7 P4 lt X g 5PX gt 4 F5 7F4gt 7 17F4 173051 33 Percentiles of Distributions The problems above concern nding probabilities that a random variable takes on certain values eg nd the probability that a certain tennis player s serve Will be over 125 miles per hour We now consider the re verse77 problem of nding values of a random variable that are obtainable With a certain probabilityr For example nd the speed at Which the tennis player can serve at least 90 of the time That is if X is the speed of the player s serve then the speed I at Which the tennis player can serve at least 90 of the time is the smallest value of x such that Fz PX S I 2 1901 This value is called the 90th percentile of X or of F In general the 100pth percentile of a random variable X or of its distribution is the smallest value of I such that Fz PX S I 2 pl 314 When X is continuous this inequality reduces to an equality because Fz is continuous in 11 The 50th percentile is called the median of X1 In addition to the median one may be interested in the 75th 90th 95th or 99th percentiles Problem 15 Finding Percentiles of a Distribution For a random variable with a given probability distribution nd its 100pth percentile by nding the smallest I that satis es Example 151 Consider the random variable X in Example 132 repre senting the number of hospitalized or dead persons in a headon collision From the table of its distribution one can see for example that its 90th percentile is 3 Of course 3 is also the 91th or 92th percentile Similarly the 99th percentile is 61 26 3 Random Variables and Their Distribution Functions Example 152 Consider the continuous random variable X in Exam ple 141 with distribution 1 7 43 Fz 3 z 6i Find its 90th percentile Solution Since X is continuous its 90th percentile is the value of X that satis es 1 7 63705 l 7 e 3 7 Solving this equation by a hand calculator yields 1 4 93 Alternatively the equation after a few manipulations is the same as Fz 90 AH 717917 e73 Taking the natural logarithm of both sides of this equation yields 3 7 z 71n17i917 3 Therefore the 90th percentile is z 7 3 711417963 7 493 Note that the same derivation applies with 9 replaced by a general probability p Therefore The 100pth percentile of X 3 7 lnl 7 10 pe S In particular for p 5 we get Median of X 3 7 lni5 i5e 3 364 34 Means and Variances of Distributions The probability distribution of a random variable contains all the detailed information about it For some purposes however it is adequate to sum marize this information by the mean and variance of the random variable The mean or expected value of a random variable X is de ned by EX if X is discrete EX dz if X is continuous It is sometimes customary to denote the mean EX by the greek letter M The mean is viewed as a onenumber macro measure of Xi lts importance 34 Means and Variances of Distributions 27 stems from the following property Suppose that one repeats the experiment associated with X several times say n times n 2 25 would be preferable and observes the values 11 r r In for Xi Then the average of these values is n 1 g E 161 This sample average of X would be very close to the mean of Xi This is why the word average is sometimes used loosely in referring to the mean For instance suppose that X denotes the number of airline tickets that a certain agent sells in one day During the last 10 days the number of tickets the agent sold was 6 5 7 2 4 6 7 3 4 4 Then the average number sold in these n 10 days is 1 10 7 21k 4s 10 kl This value would be close to EXr Note that this information was obtained without knowing anything about the distribution of Xi Suppose that X is a random variable with mean a EXr The variance of X is de ned by VarX 7 n2pz if X is discrete VarX I 7 n2fz dz if X is continuous The standard convention is to write a2 VarXi The square root 0 of the variance is called the standard deviation The variance is a macro mea sure of the spread of the probability distribution away from the mean Alternative formulas that are easier for computations are VarX 12p1 7 2 if X is discrete VarX dz 7 2 if X is continuous ln physics or mechanics the mean is called the rst moment of the distribution about the value 0 and the variance is the second moment of the distribution about a We will consider means and variances only when they are nite there are random variables whose means and variances are in nite or don7t exist Problem 16 Mean and Variance of a Random Variable Find the mean and variance of a random variable from its probability function or density by the formulas above 28 3 Random Variables and Their Distribution Functions The next two examples show that nding a mean and variance is rather mechanical in that it only involves computing sums or integrals Example 161 Consider the number of hurricanes X that hit the east coast of the United States in one yeari Suppose that its probability function is as follows I 0 1 2 3 4 5 6 7 101 0515 122 126 114 108 107103 The expected number of hurricanes in a year is EX 01051115 22 3126 4114 5108 6107 7103 2961 Thus one would expect about 3 hurricanes per yeari For the variance 2 12101 0210512115 22122 32126 42114 52108 62107 72103 1139 Then VarX 11139 7 296V 4154 Example 162 Suppose that X is a continuous random variable with den sity function 312 OS I S l and 0 elsewhere lts expected value is 00 1 EX zfzdz3 zsdz341 700 0 Its variance is 00 1 VarX z2fzdziu23 z4dzi3423801 oo 0 We now discuss an important practical use of the variance Here is a motivating problemi Suppose that the lifetime of an automobile battery is a random variable X with an unknown probability distribution but it is known that its mean is M 5 years and its standard deviation is a 12 years Using only M and a can we nd a value 1 such that at least 90 of battery lifetimes will be in the time interval 11 7 11 1 That is Pg 7 v S X S M v 2 1901 In other words the lifetime of a typical battery is 5 years plus or minus 1 years with a 90 assurance We will solve such problems by using the following Chebyshev inequality PMikaSXSMkaZlik 2r