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)