If the probability changes from trial to trial , one of the assumptions of Binomial distribution gets violated and Binomial distribution can not be used.
In such cases hypergeometric distribution is used.
This is particularly used in cases of sampling without replacement from a finite population.
For N objects of which r of type 1 and N-r of type 2 hypergeometric distribution can be given by
\[P(X=x)=\frac{(^rC_x)(^{N-r}C_{n-x})}{^NC_n}\]
Boundary conditions
N> n
r > x
N-r > n-x
Example - There are 10 markers on a table , of which 6 are defective and 4 are not defective . If 3 are randomly taken from above lot , What is probability that exactly one marker is defective?
Solution
Here r=2
x=1 Probability of being 1 marker defective
N=10 is the number of trials
n=2
by putting these values we get p(X=1)= 0.3556
Example 2
There are 25 calculators in a box. Two of them are defective. Suppose 5 calculators are randomly picked for inspection. what is probability that only one of the defective calculators will be included in the inspection.
solution
Here r=2
x=1
N=25
n=2
by putting these values in the above equation we get 1 defective in 5 calculator as 0.33
