Comparison of the Poisson and Binomial Distributions
4.6 H YPERGEOMETRIC D ISTRIBUTION
4.6 H
YPERGEOMETRICD
ISTRIBUTIONThe binomial distribution presented in Section 4.4 assumes that the items are drawn independently, with the probability of selecting an item being constant. In many ap-plied problems these assumptions can be met if a small sample is drawn from a large population. But here we consider, for example, a situation where it is necessary to select 5 employees from a group of 15 equally qualified applicants—a small population. In the group of 15 there are 9 women and 6 men. Suppose that, in the group of 5 selected employees, 3 are men and 2 are women. What is the probability of selecting that particu-lar group if the selections are made randomly without bias. In the initial group of 15, the probability of selecting a woman is 9/15. If a woman is not selected in the first drawing, then the probability of selecting a woman in the second drawing is 9/14. Thus, the prob-abilities change with each selection. Because the assumptions for the binomial are not met, a different probability must be selected. This probability distribution is the hypergeometric distribution. The hypergeometric probability distribution is given in Equation 4.23.
4.52 Determine the probability of fewer than 6 successes for a random variable with a Poisson distribution with parameter l = 3.4.
4.53 Determine the probability of fewer than or equal to 9 successes for a random variable with a Poisson distri-bution with parameter l = 8.0.
Application Exercises
4.54 Customers arrive at a busy checkout counter at an av-erage rate of 3 per minute. If the distribution of arriv-als is Poisson, find the probability that in any given minute there will be 2 or fewer arrivals.
4.55 The number of accidents in a production facility has a Poisson distribution with a mean of 2.6 per month.
a. For a given month what is the probability there will be fewer than 2 accidents?
b. For a given month what is the probability there will be more than 3 accidents?
4.56 A customer service center in India receives, on aver-age, 4.2 telephone calls per minute. If the distribution of calls is Poisson, what is the probability of receiving at least 3 calls during a particular minute?
4.57 Records indicate that, on average, 3.2 breakdowns per day occur on an urban highway during the morning rush hour. Assume that the distribution is Poisson.
a. Find the probability that on any given day there will be fewer than 2 breakdowns on this highway during the morning rush hour.
b. Find the probability that on any given day there will be more than 4 breakdowns on this highway during the morning rush hour.
4.58 Blue Cross Health Insurance reported that 4.5% of claims forms submitted for payment after a com-plex surgical procedure contain errors. If 100 of these forms are chosen at random, what is the prob-ability that fewer than 3 of them contain errors?
Use the Poisson approximation to the binomial distribution.
4.59 A corporation has 250 personal computers. The prob-ability that any 1 of them will require repair in a given week is 0.01. Find the probability that fewer than 4 of the personal computers will require repair in a partic-ular week. Use the Poisson approximation to the bino-mial distribution.
4.60 An insurance company holds fraud insurance policies on 6,000 firms. In any given year the probability that any single policy will result in a claim is 0.001. Find the probability that at least 3 claims are made in a given year. Use the Poisson approximation to the bi-nomial distribution.
4.61 A state has a law requiring motorists to carry insur-ance. It was estimated that, despite this law, 6.0% of all motorists in the state are uninsured. A random sample of 100 motorists was taken. Use the Poisson approximation to the binomial distribution to esti-mate the probability that at least 3 of the motorists in this sample are uninsured. Also indicate what calculations would be needed to find this probabil-ity exactly if the Poisson approximation was not used.
4.62 A new warehouse is being designed and a deci-sion concerning the number of loading docks is required. There are two models based on truck-arrival assumptions for the use of this warehouse, given that loading a truck requires 1 hour. Using the first model, we assume that the warehouse could be serviced by one of the many thousands of inde-pendent truckers who arrive randomly to obtain a load for delivery. It is known that, on average, 1 of these trucks would arrive each hour. For the second model, assume that the company hires a fleet of 10 trucks that are assigned full time to shipments from this warehouse. Under that assumption the trucks would arrive randomly, but the probability of any truck arriving during a given hour is 0.1. Obtain the appropriate probability distribution for each of these assumptions and compare the results.
174 Chapter 4 Discrete Probability Distributions
The preceding example describes a situation of sampling without replacement since an item drawn from the small population is not replaced before the second item is se-lected. Thus the probability of selection changes after each succeeding selection. This change is particularly important when the population is small relative to the size of the sample.
We can use the binomial distribution in situations that are defined as sampling with re-placement. If the selected item is replaced in the population, then the probability of select-ing that type of item remains the same and the binomial assumptions are met. In contrast, if the items are not replaced—sampling without replacement—the probabilities change with each selection, and, thus, the appropriate probability model is the hypergeometric distribu-tion. If the population is large 1N 7 10, 0002 and the sample size is small 16 1%2, then the change in probability after each draw is very small. In those situations the binomial is a very good approximation and is typically used.
Hypergeometric Distribution
Suppose that a random sample of n objects is chosen from a group of N objects, S of which are successes. The distribution of the number of suc-cesses, X, in the sample is called the hypergeometric distribution. Its probability distribution is
P1x2 = Csx CN-sn-x
CNn
= S!
x!1S - x2! * 1N - S2!
1n - x2!1N - S - n + x2!
N!
n!1N - n2!
(4.23)
where x can take integer values ranging from the larger of 0 and 3n - 1N - S24 to the smaller of n and S.
The logic for the hypergeometric distribution was developed in Section 3.2 using the classic definition of probability and the counting formulas for combinations. In Equation 4.23 the individual components are as follows:
1. The number of possible ways that x successes can be selected for the sample out of S successes contained in the population:
Csx = S!
x!1S - x2!
2. The number of possible ways that n - x nonsuccesses can be selected from the popu-lation that contains N - S nonsuccesses:
CN-Sn-x = 1N - S2!
1n - x2!1N - S - n + x2!
3. And, finally, the total number of different samples of size n that can be obtained from a population of size N:
CNn = N!
n!1N - n2!
When these components are combined using the classical definition of probability, the hypergeometric distribution is obtained.
The hypergeometric distribution is used for situations similar to the binomial with the important exception that sample observations are not replaced in the population when sampling from a “small population.” Therefore, the probability, P, of a success is not con-stant from one observation to the next.
Exercises 175 Hypergeometric probabilities can also be computed using computer packages fol-lowing a procedure similar to the procedure in Example 4.9 for the binomial. We would strongly recommend that you use computer computation for hypergeometric probabili-ties because using the equations is very time consuming and easily subject to errors.
Example 4.14 Shipment of Items (Compute Hypergeometric Probability)
A company receives a shipment of 20 items. Because inspection of each individual item is expensive, it has a policy of checking a random sample of 6 items from such a shipment, and if no more than 1 sampled item is defective, the remainder will not be checked. What is the probability that a shipment of 5 defective items will not be sub-jected to additional checking?
Solution If “defective” is identified with “success” in this example, the shipment contains N = 20 items and S = 5 of the 20 that are successes. A sample of n = 6 items is selected. Then the number of successes, X, in the sample has a hypergeometric distribution with the probability distribution
P1x2 = CSxCNn-x-S
CNn = C5xC156-x C206 =
5!
x!15 - x2! * 15!
1 6 - x2!19 + x2!
20!
6!14!
The shipment is not checked further if the sample contains either 0 or 1 success (defective), so that the probability of its acceptance is as follows:
P1shipment accepted2 = P102 + P112 The probability of no defectives in the sample is as follows:
P102 = 5!
0!5! * 15!
6!9!
20!
6!14!
= 0.129
The probability of 1 defective item in the sample is as follows:
P112 = 5!
1!4! * 15!
5!10!
20!
6!14!
= 0.387
Therefore, we find that the probability that the shipment of 20 items containing 5 defectives is not checked further is P1shipment accepted2 = P102 + P112 = 0.129 + 0.387 = 0.516. This is a high error rate, which indicates a need for a new accep-tance rule that requires total inspection if one or more defectives are found. With this new rule, only 12.9% of these shipments would be missed.
Basic Exercises
4.63 Compute the probability of 7 successes in a random sample of size n = 14 obtained from a population of size N = 30 that contains 15 successes.
4.64 Compute the probability of 9 successes in a random sample of size n = 20 obtained from a population of size N = 80 that contains 42 successes.
4.65 Compute the probability of 3 successes in a random sample of size n = 5 obtained from a population of size N = 40 that contains 25 successes.
4.66 Compute the probability of 8 successes in a random sample of size n = 15 obtained from a population of size N = 100 that contains 50 successes.
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XERCISES176 Chapter 4 Discrete Probability Distributions