Applied Statistics and Probability for Engineers

84 CHAPTER 3 DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS

A set of Nobjects contains

Kobjects classified as successes

N Kobjects classified as failures

A sample of size nobjects is selected randomly (without replacement) from the N

objects, where and.

Let the random variable Xdenote the number of successes in the sample. Then

Xis a hypergeometric random variableand

f 1 x 2  (3-13)

a

K

x

b a

NK

nx

b

a

N

n

b

xmax 5 0, n KN 6 to min 5 K, n 6

KN nN

Definition

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 Aand Bdenote the events that the first and second parts are nonconforming, re-

spectively. In Chapter 2, we found and. 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 Xequal the number of nonconforming parts in the sample. Then

does not, or the first part selected does not and the second part

selected conforms)

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.

P 1 X 22 P 1 both parts do not conform 2  (^150)  (^8502149)  8492 0.003
 (^1800)  (^8502150)  (^8492 150)  (^85021800)  8492 0.111
P 1 X 12 P 1 first part selected conforms and the second part selected
P 1 X 02 P 1 both parts conform 2  (^1800)  (^85021799)  8492 0.886
P 1 BƒA 2  (^49)  849 P 1 A 2  (^50)  850
The expression min is used in the definition of the range of Xbecause the maximum
number of successes that can occur in the sample is the smaller of the sample size, n,
5 K, n 6
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