The mean and the variance of a binomial distribution are 4 and 2 respectively.then, the probabitly of 2 , successes is

Options

A. `28//256`
B. `219//256`
C. `128//256`
D. `37//256`

Answer

A

Explanation

Let n and p be the parameters for the distribution.
We have,
Mean `=4` and Variance `=2`
`rArr np=4 and npq=2rArr p=q=(1)/(2)and n=8`
Let X denote the number of successes. Then,
`P(X=r)=.^8C_r((1)/(2))^r((1)/(2))^(8-1)=.^8C_(r)((1)/(2))^(8)`
`therefore` Required Probability `=P(X=2)=.^8C_2((1)/(2))^8=(28)/(256)`