Page 106 - Applied Statistics And Probability For Engineers

PQ220 6234F.Ch 03 13/04/2002 03:19 PM Page 84

84 CHAPTER 3 DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS

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. That is, selected units are not replaced before the next selection is
made. Let A and B denote the events that the first and second parts are nonconforming, re-
spectively. In Chapter 2, we found P1B ƒ A2 49 849 and P1A2 50 850 . Consequently,
knowledge that the first part is nonconforming suggests that it is less likely that the second
part selected is nonconforming.
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 noncon-
forming parts in the sample is a binomial random variable.
Let X equal the number of nonconforming parts in the sample. Then

P1X 02 P1both parts conform2 1800 85021799 8492 0.886
P1X 12 P1first part selected conforms and the second part selected
does not, or the first part selected does not and the second part
selected conforms)
1800 8502150 8492 150 85021800 8492 0.111
P1X 22 P1both parts do not conform2 150 8502149 8492 0.003

As in this example, samples are often selected without replacement. Although probabili-
ties can be determined by the reasoning used in the example above, a general formula for
computing probabilities when samples are selected without replacement is quite useful. The
counting rules presented in Section 2-1.4, part of the CD material for Chapter 2, can be used
to justify the formula given below.

Definition
A set of N objects contains
K objects classified as successes
N K objects classified as failures
A sample of size n objects is selected randomly (without replacement) from the N
objects, where K N and n N .
Let the random variable X denote the number of successes in the sample. Then
X is a hypergeometric random variable and

K N K
a b a b
x n x
f 1x2 x max50, n K N6 to min5K, n6 (3-13)
N
a b
n

The expression min5K, n6 is used in the definition 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,