HYPERGEOMETRIC DISTRIBUTION - Quantitative Techniques for management

The hypergeometric distributionarises when a random selection (without repetition) is made among objects oftwo distinct types. Typical examples:

  • Choose a team of 8 from a group of 10 boys and 7 girls
  • Choose a committee of five from the legislature consisting of 52 Democrats and 48 Republicans

Letthere be a finite population of size N, where each item can be classified aseither a success or a failure. Let there be k successes in the population. If arandom sample of size n is taken from this population, then the probability ofr successes is given by

HYPERGEOMETRIC DISTRIBUTION

Example: A retailer has 10 identical televisionsets of a company out which 4 are defective. If 3 televisions are selected atrandom, construct the probability distribution of the number of defectivetelevision sets.

Solution: Let the random variable r denote thenumber of defective televisions. In terms of notations, we can write N = 10, k= 4 and n = 3.

Thus,we can write
P(r)= 4Cr* 6C3-r/ 10C3, r= 0,1,2,3

Thedistribution of r is hypergeometric. This distribution can also be written in atabular form as given below :

hypergeometric distribution

Binomial Approximation to Hypergeometric Distribution

The Hypergeometric distribution recognises the fact that we are samplingfrom a finite population without replacement, so that the result of a sample isdependent on the samples that have gone before it. Now imagine that thepopulation is very large, so that removing a sample of size n has nodiscernible effect on the population.

Then the probability that an individualsample will have the characteristic of interest is essentially constant and hasthe value D/M, because the probability of resampling an item in thepopulation, were one to replace items after sampling, would be very small. Insuch cases, the Hypergeometric distribution can be approximated by a Binomialas follows:

Hypergeometric(n, D, M) >> Binomial(n,D/M)

The rule most often quoted is that this approximation works well when n< 0.1 M.

Example : Thereare 200 identical radios out of which 80 are defective. If 5 radios areselected at random, construct the probability distribution of the number ofdefective radios by using (i) hypergeometric distribution and (ii) binomialdistribution.

Solution:

(i) It is given that N = 200, k = 80 and n = 5.
Let r be a hypergeometric random variable which denotes the number of defective radios,then

P(r)= 80Cr* 120C5-r/ 200C5, r=0,1,2,3,4,5

The probabilities for various values of r are given in the following table:

probabilities for various values of r

(ii) To use binomial distribution, we find p = 80/200 = 0.4.

P(r) 5Cr(0.4)r (0.6) 5 r .r0,1, 2, 3,4,5

The probabilities forvarious values of r are given in the following table:

probabilities forvarious values of r

We note that these probabilities are in close conformity with the hypergeometric probabilities.


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