The mean and standard deviation of a binomial variate X are 4 and 3 respectively. Then, `P(Xge1)` is equal to

Options

A. `1-(1/4)^(16)`
B. `1-(3/4)^(16)`
C. `1-(2/3)^(16)`
D. `1-(1/3)^(16)`

Answer

B

Explanation

Mean `=n p=4` variance
`=nqp=3`
On Solving we get
`q=3/4,n=16,p=1/4`
Now `P(Xge1)=-1-P(X=0)`
`=1-.^(n)C_(0)P^(0)n^(n-0)`
`=1-(3/4)^(16)`