Introduction to Probability and Statistics for Engineers and Scientists

156 Chapter 5: Special Random Variables

Starting withP{X= 0 }=e−λ, we can use Equation 5.2.7 to successively compute

P{X= 1 }=λP{X= 0 }

P{X= 2 }=

λ

2

P{X= 1 }

P{X=i+ 1 }=

λ

i+ 1

P{X=i}

The text disk includes a program that uses Equation 5.2.7 to compute Poisson probabilities.

5.3The Hypergeometric Random Variable

A bin containsN+Mbatteries, of whichNare of acceptable quality and the otherMare
defective. A sample of sizenis to be randomly chosen (without replacements) in the sense
that the set of sampled batteries is equally likely to be any of the

(N+M

n

)
subsets of sizen.
If we letXdenote the number of acceptable batteries in the sample, then

P{X=i}=

(N

i

)(M

n−i

)

(N+M

n

), i=0, 1,..., min(N,n)∗ (5.3.1)

Any random variableXwhose probability mass function is given by Equation 5.3.1 is said
to be ahypergeometricrandom variable with parametersN,M,n.

EXAMPLE 5.3a The components of a 6-component system are to be randomly chosen from
a bin of 20 used components. The resulting system will be functional if at least 4 of its
6 components are in working condition. If 15 of the 20 components in the bin are in
working condition, what is the probability that the resulting system will be functional?

SOLUTION IfXis the number of working components chosen, thenXis hypergeometric
with parameters 15, 5, 6. The probability that the system will be functional is

P{X≥ 4 }=

∑^6

i= 4

P{X=i}

=

(

15

4

)(

5

2

)

+

(

15

5

)(

5

1

)

+

(

15

6

)(

5

0

)

(

20

6

)

≈.8687 ■

* We are following the convention that(mr)=0ifr>mor ifr<0.