The mean and variance of a random variable X having a binomial distribution are `4 and 2` respectively. The `P(X=1)` is

Options

A. `(1)/(4)`
B. `(1)/(32)`
C. `(1)/(16)`
D. `(1)/(8)`

Answer

B

Explanation

Let n and p be the parameters of the binomial distribution
We have , Mean =4 and variance =2
`implies np=4 and npq=2`
`implies p=1/2 =q and n=8`
`therefore ` Required probability `=P(X=1)=""^(8)C_(1)((1)/(2))^(1)((1)/(2))^(7)=(1)/(32)`