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C H A P T E R13Discrete ProbabilityDistributions andSimulationObjectivesTo demonstrate the basic ideas of discrete random variables.To introduce the concept of a probability distribution for a discrete random variable.To introduce and investigate applications of the binomial probability distribution.To show that simulation can be used to provide estimates of probability which areclose to exact solutions.To use simulation techniques to provide solutions to probability problems where anexact solution is too difficult to determine.To use coins and dice as simulation models.To introduce and use random number tables.13.1Discrete random variablesIn Chapter10the notion of the probability of an event occurring was explored,where an event was defined as any subset of a sample space.Sample spaces which were not sets of numberswere frequently encountered.For example,when a coin is tossed three times the samplespace is:ε={HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}If it is only the number of heads that is of interest,however,a simpler sample space could beused whose outcomes are numbers.Let X represent the number of heads in the three tosses ofthe coin,then the possible values of X are0,1,2,and3.Since the actual value that X will takeis the outcome of a random experiment,X is called a random variable.Mathematically arandom variable is a function that assigns a number to each outcome in the sample spaceε.343344Essential Mathematical Methods1&2CASA random variable X is said to be discrete if it can assume only a countable number ofvalues.For example,suppose two balls are selected at random from a jar containing several white(W)and black balls(B).A random variable X is defined as the number of white balls obtained in the sample.Thus X is a discrete random variable which may take one of thevalues0,1,or2.The sample space of the experiment is:S={WW,WB,BW,BB}Each outcome in the sample space corresponds to a value of X,and vice-versa.Experimental outcome V alue of XWW X=2BW X=1WB X=1BB X=0Many events can be associated with a given experiment.Some examples are:Event Sample outcomesOne white ball:X=1{WB,BW}At least one white ball:X≥1{WW,WB,BW}No white balls:X=0{BB}A probability distribution can be thought of as the theoretical description of a randomexperiment.Consider tossing a die600times–we might obtain the following results:x123456Frequency981049397108100Experimental probability 986001046009360097600108600100600Theoretically,the frequencies will be equal,regardless of the number of trials(provided they are sufficiently large).This can most easily be expressed by giving probabilities–for instance,let X be‘the outcome of tossing a die’.x123456Theoretical frequency100100100100100100Pr(X=x)161616161616The table gives the probability distribution of X.Chapter13—Discrete Probability Distributions and Simulation345The probability distribution of X,p(x)=Pr(X=x)is a function that assigns probabilities to each value of X.It can be represented by a rule,a table or a graph,and must give a probability p(x)for every value x that X can take.For any discrete probability function the following must be true:1The minimum possible value of p(x)is zero,and the maximum possible value of p(x)is1. That is:0≤p(x)≤1for every value x that X can take.2All values of p(x)in every probability distribution must sum to exactly1.To determine the probability that X lies in an interval,we add together the probabilities that X takes all values included in that interval,as shown in the following example.Consider the function:x12345Pr(X=x)2c3c4c5c6ca For what value of c is this a probability distribution?b Find Pr(3≤X≤5).Solutiona To be a probability distribution we require2c+3c+4c+5c+6c=120c=1c=1 20b Pr(3≤X≤5)=Pr(X=3)+Pr(X=4)+Pr(X=5)=420+520+620=1520=34The table shows a probability distribution with random variable X.x123456Pr(X=x)0.20.20.070.170.130.23346Essential Mathematical Methods 1&2CASGive the following probabilities:a Pr(X >4)b Pr(2<X <5)c Pr(X ≥5|X ≥3)Solutiona Pr(X >4)=Pr(X =5)+Pr(X =6)=0.13+0.23=0.36b Pr(2<X <5)=Pr(X =3)+Pr(X =4)=0.07+0.17=0.24c Pr(X ≥5|X ≥3)=Pr(X ≥5)Pr(X ≥3)(as X ≥5and X ≥3implies X ≥5)=Pr(X =5)+Pr(X =6)Pr(X =3)+Pr(X =4)+Pr(X =5)+Pr(X =6)=0.13+0.230.07+0.17+0.13+0.23=0.360.6=35The following distribution table gives the probabilities for the number of people on a carnival ride at a paticular time of day.No.of people (t )012345Pr(T =t )0.050.20.30.20.10.15Find:a Pr(T >4)b Pr(1<T <5)c Pr(T <3|T <4)Solutiona Pr(T >4)=Pr(T =5)=0.15b Pr(1<T <5)=Pr(T =2)+Pr(T =3)+Pr(T =4)=0.6c Pr(T <3|T <4)=Pr (T <3)Pr (T <4)=0.550.75=1115Exercise 13A1A random variable X can take the values x =1,2,3,4.Indicate whether or not each of thefollowing is a probability function for such a variable,and if not,give reasons:a p (1)=0.05p (2)=0.35p (3)=0.55p (4)=0.15b p (1)=0.125p (2)=0.5p (3)=0.25p (4)=0.0625Chapter13—Discrete Probability Distributions and Simulation347c p(1)=13%p(2)=69%p(3)=1%p(4)=17%d p(1)=51p(2)=12p(3)=34p(4)=3e p(1)=0.66p(2)=0.32p(3)=−0.19p(4)=0.212For each of the following write a probability statement in terms of the discrete random variable X showing the probability that:a X is equal to2b X is greater than2c X is at least2d X is less than2e X is2or moref X is more than2g X is no more than2h X is greater than or equal to2i X is less than or equal to2j X is no less than2k X is greater than2and less than53A random variable X can take the values0,1,2,3,4,5.List the set of values that X can take for each of the following probability statements:a Pr(X=2)b Pr(X>2)c Pr(X≥2)d Pr(X<2)e Pr(X≤2)f Pr(2≤X≤5)g Pr(2<X≤5)h Pr(2≤X<5)i Pr(2<X<5)4Consider the following function:x12345Pr(X=x)k2k3k4k5ka For what value of k is this a probability distribution?b Find Pr(2≤X≤4).5The number of‘no-shows’on a scheduled airlineflight has the following probability distribution:r01234567p(r)0.090.220.260.210.130.060.020.01Find the probability that:a more than four people do not show up for theflightb at least two people do not show up for theflight.6Suppose Y is a random variable with the distribution given in the table.y0.20.30.40.50.60.70.80.9 Pr(Y=y)0.080.130.090.190.200.030.100.18 Find:a Pr(Y≤0.50)b Pr(Y>0.50)c Pr(0.30≤Y≤0.80)348Essential Mathematical Methods1&2CAS7The table shows a probability distribution with random variable X.x123456Pr(X=x)0.10.130.170.270.200.13Give the following probabilities:a Pr(X>3)b Pr(3<X<6)c Pr(X≥4|X≥2)8Suppose that a fair coin is tossed three times.a List the eight equally likely outcomes.b If X represents the number of heads shown,determine Pr(X=2).c Find the probability distribution of the random variable X.d Find Pr(X≤2).e Find Pr(X≤1|X≤2).9When a pair of dice is rolled,36equally likely outcomes are possible.Let Y denote the sum of the dice.a What are the possible values of the random variable Y?b Find Pr(Y=7).c Determine the probability distribution of the random variable Y.10When a pair of dice is rolled,36equally likely outcomes are possible.Let X denote the larger of the values showing on the dice.If both dice come up the same,then X denotesthe common value.a What are the possible values of the random variable X?b Find Pr(X=4).c Determine the probability distribution of the random variable X.11A dart is thrown at a circular board with a radius of10cm.The board has three rings:a bullseye of radius3cm,a second ring with an outer radius of7cm,and a third ring withouter radius10cm.Assume that the probability of the dart hitting a region R is given byPr(R)=area of R area of dartboarda Find the probability of scoring a bullseye.b Find the probability of hitting the middle ring.c Find the probability of hitting the outer ring.12Suppose that a fair coin is tossed three times.Y ou lose$3.00if three heads appear and $2.00if two heads appear.Y ou win$1.00if one head appears and$3.00if no heads appear;$Y is the amount you win or lose.a Find the probability distribution of the random variable Y.b Find Pr(Y≤1).Chapter 13—Discrete Probability Distributions and Simulation 34913.2Sampling without replacementConsider the sort of probability distribution that arises from the most common sampling situation.A jar contains three mints and four toffees,and Bob selects two (without looking).If the random variable of interest is the number of mints he selects,then this can take values of 0,1or 2.Suppose this was done experimentally many times (say 50).The following results may be obtained:Number of mints 012Number of times observed 18248Let X be the number of mints Bob selects.From the table,the probabilities can be estimated for each outcome as:Pr(X =0)≈1850=0.36Pr(X =1)≈2450=0.48Pr(X =2)≈850=0.16Obviously,it is not desirable that an experiment needs to be carried out every time a situation like this arises and it is not necessary.It is possible to work out the theoretical probability for each value of the random variable by using the knowledge of combinations from Chapter 12.Consider the situation in which the sample of two sweets contains no mints.Then it must contain two toffees.Thus Bob has selected no mints from the three available,and two toffees from the four available,which gives the number of favourable outcomes as:30 42Now the number of possible choices Bob has of choosing two sweets from seven is 72 andthus the probability of Bob selectingno mints is:Pr(X =0)= 30 4272=621=27350Essential Mathematical Methods 1&2CASSimilarly,we can determine:Pr(X =1)= 31 41 72 and Pr(X =2)= 32 4072=1221=321=47=17Thus the probability distribution for X is:x 012Pr(X =x )274717Note that the probabilities in the table add up to 1.If they did not add to 1it would be known that an error had been made.This problem can also easily becompleted with a tree diagram.MintTherefore Pr(X =0)=47×36=27Pr(X =1)=47×36+37×46=27+27=47and Pr(X =2)=37×26=17This can be considered as a sequence of two trials in which the second is dependent on the first.Instead of evaluating the probabilities for all the values of X and listing them in a table,the probability distribution could be given as a rule and,providing the values of X are specified for which the rule is appropriate,the same information as before is available.In this case the rule is:Pr(X =x )= 3x42−x72,x =0,1,2This example is an application of a distribution which is commonly called the hypergeometric distribution .Chapter 13—Discrete Probability Distributions and Simulation351Marine biologists are studying a group of dolphins which live in a small bay.They know there are 12dolphins in the group,four of which have been caught,tagged and released to mix back into the population.If the researchers return the following week and catch another group of three dolphins,what is the probability that two of these will already be tagged?SolutionLet X equal the number of tagged dolphins in the second sample.We wish to know the probability of selecting two of the four tagged dolphins,andone of the eight non tagged dolphins,when a sample of size 3is selected from apopulation of size 12.That is:Pr(X =2)= 42 81123=1255Exercise 13B1A company employs 30salespersons,12of whom are men and 18are women.Fivesalespersons are to be selected at random to attend an important conference.What is the probability of selecting two men and three women?2An electrical component is packaged in boxes of 20.A technician randomly selects threefrom each box for testing.If there are no faulty components,the whole box is passed.If there are any faulty components,the box is sent back for further inspection.If a box is known to contain four faulty components,what is the probability it will pass?3A pond contains seven gold and eight black fish.If three fish are caught at random in a net,find the probability that at least one of them is black.4A researcher has caught,tagged and released 10birds of a particular species into the forest.If there are known to be 25of this species of bird in the area,what is the probability that another sample of five birds will contain three tagged ones?5A tennis instructor has 10new and 10used tennis balls.If he selects six balls at random touse in a class,what is the probability that there will be at least two new balls?6A jury of six persons was selected from a group of 18potential jurors,of whom eight werefemale and 10male.The jury was supposedly selected at random,but it contained only one female.Do you have any reason to doubt the randomness of the selection?Explain your reasons.352Essential Mathematical Methods 1&2CAS 13.3Sampling with replacement:the binomial distributionSuppose a fair six-sided die is rolled four times and a random variable X is defined asthenumber of 3s observed.An approximate probability distribution for this random variable could be found by repeatedly rolling the die four times,and observing the outcomes.On one occasion the four rolls of the die was repeated 100times and the following resultsnoted:x 01234No.of times observed 5044510From this table the probabilities for each outcome could be estimated:Pr(X =0)≈50100=0.50Pr(X =1)≈40100=0.44Pr(X =2)≈5100=0.05Pr(X =3)≈1100=0.01Pr(X =4)≈0100=0The theoretical probability distribution can be determined in the following way.One possible outcome of the experiment is TTNN ,where T represents a 3and N represents not a 3.The probability of this particular outcome (that is,in this order)is16×16×56×56= 16 2562How many different arrangements of T,T,N ,and N are there?Listing we find that there are six:TTNN,TNTN,TNNT,NTTN,NTNT,NNTT .The number of arrangements could be found without listing them by recognising that this is equal to 42 ,the number of ways of placingthe two T s in the four available places.Thus,the probability of obtaining exactly two 3s when a fair die is tossed four times is:Pr(X =2)=42 16 2 562=25216Continuing in this way the entire probability distribution can be defined as given in the table.(Note that the probabilities shown do not add to exactly 1owing to rounding errors.)x 01234Pr(X =x )0.48220.38580.11570.01540.0008It would be convenient to be able to use a formula to summarise the probability distribution.In this case it is:Pr(X =x )= 4x 16x 56 n −xx =0,1,2,3,4This is an example of the binomial probability distribution ,which has arisen from abinomial experiment.A binomial experiment is one that possesses the following properties:The experiment consists of a number,n ,of identical trials.Each trial results in one of two outcomes,which are usually designated as either a success ,S ,or a failure ,F .The probability of success on a single trial,p say,is constant for all trials (and thus the probability of failure on a single trial is (1–p )).The trials are independent (so that the outcome on any trial is not affected by the outcome of any previous trial).The random variable of interest,X ,is the number of successes in n trials of a binomial experiment.Thus,X has a binomial distribution and the rule is:Pr(X=x )= n x (p )x(1−p )n −x x =0,1,...,nwhere n x =n !x !(n −x )!Rainfall records for the city of Melbourne indicate that,on average,the probability of rain falling on any one day in November is 0.4.Assuming that the occurrence of rain on any day is independent of whether or not rain falls on any other day,find the probability that rain will fall on any three days of a chosen week.SolutionSince there are only two possible outcomes on each day (rain or no rain),theprobability of rain on any day is constant (0.4)regardless of previous outcomes.The situation described is a binomial experiment.In this example occurrence of rain isconsidered as a success,and so define X as the number of days on which it rains in a given week.Thus X is a binomial random variable with p =0.4and n =7.Pr(X=x)=7x(0.4)x(0.6)7−x,x=0,1,...,7and Pr(X=3)=73(0.4)3(0.6)7−3=7!3!4!×0.064×0.13=0.290304(values held in calculator)the binomialcumulativegiven in Example6.Example6For the situation described calculator tofind the probabilitywhere n=7,p=0.4.shown.Once again it is found Pr(X=3)=0.2903.Here Pr(X ≤3)=0.7102.This time Pr(X ≥3)is required:Pr(X ≥3)=Pr(X =3)++Pr(X =5)+Pr(X =6)The CAS calculator does built-in function to calculate but it can be used to evaluate probability by recognising that:Pr(X ≥3)=1−(Pr(X ==1)+Pr(X =2))=1−Pr(X Here Pr(X ≤2)=and hence Pr(X ≥3)==0.5801CAS calculator can also display thebinomial probability distribution graphically.Thisshown in Example 7.Example 7Use the CAS calculator to plot the probability distribution functionPr(X =x )= n x p x (1x =0,1,...,nentry for Binomial Pdf without in List 1and then a scatterplotExample8The probability of winningExample9The probability of an archer from a shot is0.4.Find themaximum score:maximum score at least once.shots.b Pr(X=3|X>0)=Pr(X=3) Pr(X>0)=0.23041−Pr(X=0)=0.23041−0.65=0.23040.92224=0.2498(correct to4decimal places) Exercise13C1For the binomial distribution Pr(X=x)=6x(0.3)x(0.7)6−x,x=0,1,...,6,find:a Pr(X=3)b Pr(X=4)2For the binomial distribution Pr(X=x)=10x(0.1)x(0.9)10−x,x=0,1,...,10,find:a Pr(X=2)b Pr(X≤2)3A fair die is e your CAS calculator tofind the probability of observing:a exactly ten6sb fewer than ten6sc at least ten6s4Rainfall records for the city of Melbourne indicate that,on average,the probability of rain falling on any one day in November is0.35.Assuming that the occurrence of rain on any day is independent of whether or not rain falls on any other day,find the probability that:a rain will fall on thefirst three days of a given week,but not on the other fourb rain will fall on exactly three days of a given week,but not on the other fourc rain will fall on at least three days of a given week.5A die is rolled seven times and the number of2s that occur in the seven rolls is noted.Find the probability that:a thefirst roll is a2and the rest are notb exactly one of the seven rolls results in a2.6If the probability of a female child being born is0.5,use your CAS calculator tofind the probability that,if100babies are born on a certain day,more than60of them will be female.7A breakfast cereal manufacturer places a coupon in every tenth packet of cereal entitling the buyer to a free packet of cereal.Over a period of two months a family purchasesfive packets of cereal.a Find the probability distribution of the number of coupons in thefive packets.b What is the most probable number of coupons in thefive packets?8If the probability of a female child being born is0.48,find the probability that a family with exactly three children has at least one child of each sex.9An insurance company examines its records and notes that30%of accident claims are made by drivers aged under21.If there are100accident claims in the next12months,use your CAS calculator to determine the probability that40or more of them are made by drivers aged under21.10A restaurant is able to seat80customers inside,and many more at outside tables.Generally,80%of their customers prefer to sit inside.If100customers arrive one day,use your CAS calculator to determine the probability that the restaurant will seat inside all those who make this request.11A supermarket has four checkouts.A customer in a hurry decides to leave without making a purchase if all the checkouts are busy.At that time of day the probability of each checkout being free is0.25.Assuming that whether or not a checkout is busy isindependent of any other checkout,calculate the probability that the customer will make a purchase.12An aircraft has four engines.The probability that any one of them will fail on aflight is0.003.Assuming the four engines operate independently,find the probability that on aparticularflight:a no engine failure occursb not more than one engine failure occursc all four engines fail13A market researcher wishes to determine if the public has a preference for one of two brands of cheese,brand A or brand B.In order to do this,15people are asked to choose which cheese they prefer.If there is actually no difference in preference:a What is the probability that10or more people would state a preference for brand A?b What is the probability that10or more people would state a preference for brand A orbrand B?14It has been discovered that4%of the batteries produced at a certain factory are defective.A sample of10is drawn randomly from each hour’s production and the number ofdefective batteries is noted.In what percentage of these hourly samples would there be a least two defective batteries?Explain what doubts you might have if a particular sample contained six defective batteries.15An examination consists of10multiple-choice questions.Each question has four possible answers.At leastfive correct answers are required to pass the examination.a Suppose a student guesses the answer to each question.What is the probability thestudent will make:i at least three correct guesses?ii at least four correct guesses?iii at leastfive correct guesses?b How many correct answers do you think are necessary to decide that the student is notguessing each answer?Explain your reasons.16An examination consists of20multiple-choice questions.Each question has four possible answers.At least10correct answers are required to pass the examination.Suppose the student guesses the answer to each e your CAS calculator to determine theprobability that the student passes.17Plot the probability distribution function Pr(X=x)= nxp x(1−p)n−x,x=0,1,...,n,for n=10and p=0.318Plot the probability distribution function Pr(X=x)= nxp x(1−p)n−x,x=0,1,...,n,for n=15and p=0.6.19What is the least number of times a fair coin should be tossed in order to ensure that:a the probability of observing at least one head is more than0.95?b the probability of observing more than one head is more than0.95?20What is the least number of times a fair die should be rolled in order to ensure that:a the probability of observing at least one6is more than0.9?b the probability of observing more than one6is more than0.9?21Geoff has determined that his probability of hitting an ace when serving at tennis is0.1.What is the least number of balls he must serve to ensure that:a the probability of hitting at least one ace is more than0.8?b probability of hitting more than one ace is more than0.8?22The probability of winning in a game of chance is known to be0.05.What is the least number of times Phillip should play the game in order to ensure that:a the probability that he wins at least once is more than0.90?b the probability that he wins at least once is more than0.95?23The probability of a shooter obtaining a maximum score from a shot is0.7.Find the probability that out offive shots the shooter obtains the maximum score:a three timesb three times,given that it is known that he obtains the maximum score at least once.24Each week a securityfirm transports a large sum of money between two places.The day on which the journey is made is varied at random and,in any week,each of thefive daysfrom Monday to Friday is equally likely to be chosen.(In the following,give answerscorrect to4decimal places.)Calculate the probability that in a period of10weeks Friday will be chosen:a two timesb at least two timesc exactly three times,given it is chosen at least two times13.4Solving probability problems using simulationSimulation is a very powerful and widely used procedure which enables us tofind approximate answers to difficult probability questions.It is a technique which imitates the operation of thereal-world system being investigated.Some problems are not able to be solved directly andsimulation allows a solution to be obtained where otherwise none would be possible.In thissection some specific probability problems are looked at which may be solved by usingsimulation,a valuable and legitimate tool for the statistician.What is the probability that a family offive children will include at least four girls?SolutionThis problem could be simulated by tossing a coinfive times,once for each child,using a model based on the following assumptions:There is a probability of0.5of each child being female.The sex of each child is independent of the sex of the other children.That is,theprobability of a female child is always0.5.Since the probability of a female child is0.5,then tossing a fair coin is a suitablesimulation model.Let a head represent a female child and a tail a male child.A trialconsists of tossing the coinfive times to represent one complete family offivechildren and the result of the trial is the number of female children obtained in thetrial.To estimate the required probability several trials need to be conducted.Howmany trials are needed to estimate the probability?As we have already noted inSection8.2,the more repetitions of an experiment the better the estimate of theprobability.Initially about50trials could be considered.An example of the results that might be obtained from10trials is given in the tableon the next page:Trial number Simulation results Number of heads1THHTT 22HHHTH 43HHHTH 44HTTTH 25HTHHH 46HTTTH 27TTHHH 38HTHHT 39TTTHH 210HHTTT 2Continuing in this way,the following results were obtained for 50trials:Number of heads Number of times obtained011821731341051The results in the table can be used to estimate the required probability.Since atleast four heads were obtained in 11trials,estimate the probability of at least fourfemale children as 1150or 0.22.Of course,since this probability has been estimatedexperimentally,repeating the simulations would give a slightly different result,but we would expect to obtain approximately this value most of the time.Example 10can be recognised as a situation involving a binomial random variable,with n =5and p =0.5.Thus the exact answer to the question ‘What is the probability that a family of five children will include at least four girls?’is:Pr(X ≥4)= 54(0.5)4(0.5)1+ 55 (0.5)5(0.5)0=0.1875This is reasonably close to the answer obtained from the simulation.In Example 10simulation was used to provide an estimate of the value of a particularprobability.Simulation is also widely used to estimate the values of other quantities which are of interest in a probability problem.We may wish to know the average result,the largest result,the number of trials required to achieve a certain result,and so on.An example of this type of problem is given in Example 11.。