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Section 3-8/Hypergeometric Distribution 93

3-8 Hypergeometric Distribution

In Example 3-8, a day’s production of 850 manufactured parts contains 50 parts that do not
conform to customer requirements. Two parts are selected at random without replacement
from the day’s production. Let A and B denote the events that the irst and second parts are

| (
/
/
nonconforming, respectively. In Chapter 2, we found P B A) = 49 849 and P A ( ) = 50 850.

Consequently, knowledge that the irst part is nonconforming suggests that it is less likely that
the second part selected is nonconforming.
Let X equal the number of nonconforming parts in the sample. Then,
(
P X = ) = P(both parts conform ) = 800 799 = 0 886
0
.
·
850 849
(
P X = ) = P (irst part selected conforms and the second part selected does not, or the i rst

1
part selected does not and the second part selected conforms)
= 800 50 + 50 800 = 0 111.
·
·
850 849 850 849
(
P (X = ) =2 P both parts do not cconform) = 50 · 49 = 0 003
.
850 849
This experiment is fundamentally different from the examples based on the binomial dis-
tribution. In this experiment, the trials are not independent. Note that, in the unusual case that
each unit selected is replaced before the next selection, the trials are independent and there is
a constant probability of a nonconforming part on each trial. Then the number of nonconform-
ing parts in the sample is a binomial random variable.
But as in Example 3-8, samples are often selected without replacement. Although prob-
abilities can be determined by the preceding reasoning, a general formula for computing prob-
abilities when samples are selected without replacement is quite useful. The counting rules
presented in Chapter 2 can be used to justify the following formula.
Hypergeometric
Distribution A set of N objects contains

K objects classiied as successes
N − K objects classiied as failures

A sample of size n objects is selected randomly (without replacement) from the N
objects where K ≤ N and n ≤ N.
The random variable X that equals the number of successes in the sample is a
hypergeometric random variable and
⎛ K⎞ ⎛ N − K⎞
⎜ ⎟ ⎜ n x ⎠ ⎟
−
⎝
x ⎠ ⎝
}
+
f x ( ) = x = max {0 , n K − N} to min { K, n} (3-13)
⎛ N⎞
⎜ ⎟
n⎠
⎝
{
The expression min K n} is used in the dei nition of the range of X because the maximum
,
number of successes that can occur in the sample is the smaller of the sample size, n, and the
number of successes available, K. Also, if n K+ > N, at least n K+ − N successes must occur
in the sample. Selected hypergeometric distributions are illustrated in Fig. 3-12.
Example 3-26 Sampling Without Replacement The computations at the start of this section can be reanalyzed
by using the general expression in the dei nition of a hypergeometric random variable. That is,