2. Hypergeometric Distribution
Suppose we are interested in the number of defectives in
a sample of size n units drawn from a lot containing N
units, of which a are defective.
Each object has same chance of being selected, then the
probability that the first drawing will yield a defective
unit = a/N
but for the second drawing, probability
a 1
if first unit is defective,
N 1
a
if first unit is not defective.
N 1
3. Hypergeometric Distribution
• The trials here are not independent and hence the
fourth assumption underlying the binomial
distribution is not fulfilled and therefore, we cannot
apply binomial distribution here.
• Binomial distribution would have been applied if we
do sampling with replacement, viz., if each unit
selected from the sample would have been replaced
before the next one is drawn.
4. Hypergeometric Distribution
Sampling without replacement
Number of ways in which x successes (defectives) can be
chosen is a
x
Number of ways in which n – x failures (non defectives)
be chosen is N a
n x
Hence number of ways x successes and n – x failures can
be chosen is a N a
x n x
5. Hypergeometric Distribution
Number of ways n objects can be chosen from N objects is N
If all the possibilities are equally likely then for sampling n
without replacement the probability of getting “x successes
in n trials” is given by
a N a
x n x
h ( x; n, a , N ) for x 0 , 1,...., n
N
n
where x a, n x N a.
6. Hypergeometric Distribution
• The solution of the problem of sampling without
replacement gave birth to the above distribution
which we termed as hypergeometric
distribution.
• The parameters of hypergeometric distribution are
the sample size n, the lot size (or population size)
N, and the number of “successes” in the lot a.
• When n is small compared to N, the composition
of the lot is not seriously affected by drawing the
sample and the binomial distribution with
parameters n and p = a/N will yield a good
approximation.
7. Hypergeometric Distribution
The difference between the two values is only 0.010.
In general it can be shown that
h( x; n, a, N) b( x; n, p)
with p = (a/N) when N ∞.
A good rule of thumb is to use the binomial distribution
as an approximation to the hyper-geometric
distribution if n/N ≤0.05
8. The Mean and the Variance of a Probability
Distribution
Mean of hypergeometric distribution
a n sample size
n N population size
N a number of success
Proof: a N a
n n x n x
x .h ( x ; n , a , N ) x.
N
x 0 x 1
n
9. The Mean and the Variance of a Probability
Distribution
a a! a (a 1)! a a 1
x x! ( a x )! x (x 1)! ( a x )! x x 1
a 1 N a
n n a 1 N a
x 1 n x a
a.
N N x 1 n x
x 1 x 1
n n
10. The Mean and the Variance of a Probability
Distribution
Put x – 1= y
k n 1
n 1
a a 1 N a m a 1
N y 0 y n 1 y r y
n s N a
k m s m s
Use the identity
r k r k
r 0
11. The Mean and the Variance of a Probability
Distribution
We get
a N 1
N n 1
n
a
n
N
12. Variance of hypergeometric distribution
2 n a (N a) (N n)
2
N (N 1)
Proof: a N a
n n
2 2 x n x
2 x .h ( x ; n , a , N ) x .
x 0
N
x 1
n
13. a 1 N a
n x 1 n x
2 a x.
N
x 1
n
n a 1 N a
a
(x 1 1).
N x 1 n x
x 1
n
14. n a 2 N a
a (a 1)
2 .
N x 2 n x
x 2
n
n a 1 N a
a
N x 1 n x
x 1
n
Put x – 2 = y in 1st summation and x – 1 = z in 2nd one
15. n 2 a 2 N a
a (a 1)
2 .
N y n 2 y
y 0
n
n 1 a 1 N a
a
N z n 1 z
z 0
n
k m s m s
Use the identity
r k r k
r 0
k n 2, m a 2 k n 1, m a 1
r y, s N a r z, s N a
16. a (a 1) N 2 a N 1
2
N n 2 N n 1
n n
n(n 1) n
a (a 1) a
N (N 1) N
2 2 n a (N a) (N n)
2 2
N (N 1)
