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

(c ) 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), `" " nxx(1)/(2)=2`
`:. " " n=4`
The binomial distribution is `((1)/(2)+(1)/(2))^(4)`
Now, `P(X ge 1)=P(X=2)+P(X=3)+P(X=4)`
`=""^(4)C_(r )((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)`