The mean and the variance in a bionomial distribution of found to be 2 and 1 respectively. The probability P(X=0) is

Options

A. `(1)/(2)`
B. `(1)/(4)`
C. `(1)/(8)`
D. `(1)/(16)`

Answer

D

Explanation

np=2
n Pq=1
`q=(1)/(2)`
p+q=1
`P=1-(1)/(2)=(1)/(2)`
`P=1-(1)/(2)=(1)/(2)`
`n((1)/(2))=2`
`therefore n=4`
`.^(4)C_(0)=((1)/(2))^(4-0)=(1)/(16)`