Introduction to Probability and Statistics for Engineers and Scientists

5.3The Hypergeometric Random Variable 157

To compute the mean and variance of a hypergeometric random variable whose prob-
ability mass function is given by Equation 5.3.1, imagine that the batteries are drawn
sequentially and let

Xi=

{

1 if theith selection is acceptable

0 otherwise

Now, since theith selection is equally likely to be any of theN+Mbatteries, of which
Nare acceptable, it follows that

P{Xi= 1 }=

N

N+M

(5.3.2)

Also, fori=j,

P{Xi=1,Xj= 1 }=P{Xi= 1 }P{Xj= 1 |Xi= 1 }

=

N

N+M

N− 1

N+M− 1

(5.3.3)

which follows since, given that theith selection is acceptable, thejth selection is equally
likely to be any of the otherN+M−1 batteries of whichN−1 are acceptable.
To compute the mean and variance ofX, the number of acceptable batteries in the
sample of sizen, use the representation

X=

∑n

i= 1

Xi

This gives

E[X]=

∑n

i= 1

E[Xi]=

∑n

i= 1

P{Xi= 1 }=

nN

N+M

(5.3.4)

Also, Corollary 4.7.3 for the variance of a sum of random variables gives

Var(X)=

∑n

i= 1

Var(Xi)+ 2

∑∑

1 ≤i<j≤n

Cov(Xi,Xj) (5.3.5)

Now,Xiis a Bernoulli random variable and so

Var(Xi)=P{Xi= 1 }(1−P{Xi= 1 })=

N

N+M

M

N+M

(5.3.6)