6.4 THE HYPERGEOMETRIC PROBABILITY …

1 Section The HYPERGEOMETRIC PROBABILITY Distribution6 THE HYPERGEOMETRIC PROBABILITYDISTRIBUTIONP reparing for This SectionBefore getting started, review the following:Objectives1 Determine whether a PROBABILITY experiment is a hypergeometricexperiment2 Compute the probabilities of HYPERGEOMETRIC experiments3 Compute the mean and standard deviation of a HYPERGEOMETRIC random variable1 Determine Whether a ProbabilityExperiment Is a HypergeometricExperimentIn Section , we presented binomial experiments. Recall, the binomial probabilitydistribution can be used to compute the probabilities of experiments when there area fixed number of trials in which there are two mutually exclusive outcomes and theprobability of success for any trial is constant. In addition, the trials must be inde-pendent.

2 Based on the results from Example 6 in Section , we learned that, whensmall samples are obtained from large finite populations, it is reasonable to assumeindependence of events. That is, when obtaining a sample of size nfrom a populationwhose size is N, we are willing to assume independence of the events provided that(the sample size is less than 5% of the population size).What if the requirement of independence is not satisfied? Under these circum-stances, the experiment is a HYPERGEOMETRIC for a HYPERGEOMETRIC PROBABILITY ExperimentA PROBABILITY experiment is said to be a HYPERGEOMETRIC finite population to be sampled has each trial of the experiment, there are two possible outcomes,successor failure. There are exactly ksuccesses in the sample of size nis obtained from the population of size a PROBABILITY experiment satisfies these three requirements, the randomvariable X, the number of successes in ntrials of the experiment, follows thehypergeometric PROBABILITY distribution.

3 We now introduce the notation that we will Used in the HYPERGEOMETRIC PROBABILITY Distribution The population is size N. The sample is size n. There are ksuccesses in the population. Let the random variable Xdenote the number of successes in the sample ofsize n, so xmust be greater than or equal to the larger of 0 or and xmust be less than or equal to the smaller of nor , Classical Method (Section , pp. 263 266) Independence (Section , pp. 286 287) Multiplication Rule (Example 6, Section , p. 297) Multiplication Rule of Counting (Section ,pp. 302 305) Combinations (Section , pp. 307 309) 7/30/08 6:36 PM Page 6 1 Copyright 2010 Pearson Education, 1A HYPERGEOMETRIC PROBABILITY ExperimentProblem:Suppose that a researcher goes to a small college with 200 faculty, 12 ofwhich have blood type O-negative.

4 She obtains a simple random sample of of the faculty and finds that 3 of the faculty have blood type O-negative. Is thisexperiment a HYPERGEOMETRIC PROBABILITY experiment? List the possible values ofthe random variable X, the number of faculty that have blood type :We need to determine if the three criteria for a HYPERGEOMETRIC exper-iment have been :This is a HYPERGEOMETRIC PROBABILITY experiment population consists of outcomes are possible: the faculty member has blood type O-negative or thefaculty member does not have blood type O-negative. The researcher sample is size The possible values of the random variable are The largest valueof Xis 12, because we cannot have more than 12 successes since there are only12 faculty with blood type O-negative in the that we cannot use the binomial PROBABILITY distribution to determinethe likelihood of obtaining three successes in 20 trials in Example 1 because thesample size is large relative to the population size.

5 That is,is more than 5%of the population size,N= , 1, 2, , 2 Chapter 6 Discrete PROBABILITY DistributionsThe logic behind Formula (1) is based on the Classical Method given on page263, along with the Multiplication Rule of Counting given on page 304. The Classi-cal Method for computing probabilities states that the PROBABILITY of an event is thenumber of ways the event can occur, divided by the total number of outcomes inHistorical NoteThe name hypergeometricis attributed to Leonhard Euler. Eulerwas born in Basel, Switzerland, onApril 15, 1707. His father was aminister and wanted Leonhard tostudy theology as well. However, aftera discussion with Johann Bernoulli, afriend from college, Euler s fatherallowed him to study mathematicsat the University of Basel.

6 Eulercompleted his studies in 1726. Eulermarried Katharina Gsell on January 7,1734. They had 13 children, only 5 ofwhom survived. Euler claims to havemade many of his greatest discoverieswith a child in his arms and childrencrawling at his feet. In 1740, Euler lostsight in his right eye. One of his famousquotes on this loss is Now I will haveless distraction. He eventually lostsight in his other eye as well, but thisdid not slow him down. Euler died onSeptember 18, 1783, in St. Work Problem 5 2 Compute the Probabilities ofHypergeometric ExperimentsThe basis for computing probabilities in a HYPERGEOMETRIC experiment lies in thefact that each sample of size nis equally likely to be chosen. Consider an urn thatcontains 8 white chips and 6 black chips for a total of If we decide torandomly select all possible combinations of chips are equally is, if we let W1,W2,W8represent the 8 white chips and B1,B2,B6rep-resent the 6 black chips, selecting W1,W2,B3is just as likely as selecting W3,W6, in both cases that we selected 2 white chips and 1 black chip.

7 So, if Xrepre-sents the number of black chips selected, we have in both cases; however, thechips selected are different (so each represents a different sample).x=1 , ,n=3 chips,N=14 PROBABILITY DistributionThe PROBABILITY of obtaining xsuccesses based on a random sample of size nfrom a population of size Nis given by(1)where kis the number of successes in the 1kCx21N-kCn-x2 NCn 7/30/08 6:36 PM Page 6 2 Copyright 2010 Pearson Education, The HYPERGEOMETRIC PROBABILITY Distribution6 3the experiment. The denominator of Formula (1) represents the number of ways nobjects can be selected from Nobjects. This represents the number of possible out-comes in the experiment. The numerator consists of two factors. The first factor,, represents the number of ways we can select the xsuccesses from the ksuc-cesses in the population.

8 The second factor,represents the number ofways we can select failures from the failures in the population. Usingthe Multiplication Rule of Counting, we find the number of ways we could obtain xsuccesses from ntrials of the 1N-k2C1n-x2, kCxEXAMPLE 2 Using the HYPERGEOMETRIC PROBABILITY DistributionProblem:Suppose a researcher goes to a small college of 200 faculty, 12 of whichhave blood type O-negative. She obtains a simple random sample of of thefaculty. Let the random variable Xrepresent the number of faculty in the sample ofsize that have blood type O-negative.(a)What is the PROBABILITY that 3 of the faculty have blood type O-negative?(b)What is the PROBABILITY that at least one of the faculty has blood type O-negative?Approach:This is a HYPERGEOMETRIC experiment with andThe possible values of the random variable Xare (Oursample cannot have more than faculty with blood type O-negative.)

9 We useFormula (1) to compute the (a)We are looking for the PROBABILITY of obtaining 3 successes, so There is a PROBABILITY that, in a random sample of 20 faculty, exactly 3 haveblood type O-negative. If we conducted this experiment 100 times, we would expectto select 3 faculty that have blood type O-negative about 8 times.(b)The phrase at leastmeans greater than or equal to. The values of the randomvariable Xthat are greater than or equal to 1 are Computing proba-bilities for all these random variables is time consuming. It is much easier to use theComplement Rule and compute There is a PROBABILITY that, in a random sample of 20 faculty, at least 1 hasblood type O-negative. If we conducted this experiment 100 times, we would expectto select at least one of the faculty that have blood type O-negative about 73 12=1-P102=1-112C021200-12C20-02 200C20 =1-112C021188C202 200C20 = 12= , 2, 3, , 20-32 200C20 = 112C321188C172 200C20 = , 1, 2, , , n=20,n=20n=20 EXAMPLE 3 Using the HYPERGEOMETRIC PROBABILITY DistributionProblem:The HYPERGEOMETRIC PROBABILITY distribution is used in acceptance sam-pling.

10 Suppose that a machine shop orders 500 bolts from a supplier. To determinewhether to accept the shipment of bolts, the manager of the facility randomly selects12 bolts. If none of the 12 randomly selected bolts is found to be defective, he con-cludes that the shipment is acceptable.(a)If 10% of the bolts in the population are defective, what is the PROBABILITY thatnone of the selected bolts are defective?(b)If 20% of the bolts in the population are defective, what is the PROBABILITY thatnone of the selected bolts are defective? 7/30/08 6:36 PM Page 6 3 Copyright 2010 Pearson Education, 4 Chapter 6 Discrete PROBABILITY DistributionsApproach:This is a HYPERGEOMETRIC experiment with and Inpart (a), we have that defectives. The possible values of the ran-dom variable Xare (you cannot have more successes than thesample size).