If the mean and the variance of a binomial variable X are 2 and 1 respectively, then the probability that X takes a value greater than one is equal to:

Options

A. `5/16`
B. `8/16`
C. `11/16`
D. `1/16`

Answer

C

Explanation

Given mean `np=2`………..i
and variance`npq=1`…………..ii
From Eqs (i) and (ii) we get
`q=1/2`
`:.p=1-q=1/2`
From Eq. (i)
`nxx1/2=2`
`:.n=4`
The binomial distribution is
`(1/2+1/2)^(4)`
Now `P(Xgtq)=P(X=2)`
`+P(X=3)+P(X=4)`
`=.^(4)C_(2)(1/2)^(2)(1/2)^(2)+.^(4)C_(3)(1/2)^(3)(1/2)^(1)+.^(4)C_(4)(1/2)^(4)`
`=(6+4+1)/16=11/16`