Question

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

• A. $11/16$
• B. $5/16$
• C. $9/16$
• D. $7/16$

Answer 1

The correct option is A $11/16$

Here, $np=2,npq=1$
$\therefore p=q=\frac{1}{2},n=4$
Therefore $P(X>1)=1-P(X=0)-P(X=1)$
$=1{-}^4{C}_0{\left(\frac{1}{2}\right)}^4{-}^4{C}_1{\left(\frac{1}{2}\right)}^4=\frac{11}{16}$