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

$\frac{15}{16}$

Let n and p be the parameters of the binomial distribution. Then,

$np=2 $ and $ npq= 1$

$⇒ q=\frac{1}{2},p=\frac{1}{2}$ and n = 4           [Given]

$∴ P(X ≥ 1)= 1- P(X<1) = 1- P(X= 0)$

$⇒P(X ≥ 1)= 1 - {^nC}_0 \, p^0q^{n-0}= 1- {^4C}_0 × \left(\frac{1}{2}\right)^4=\frac{15}{16}$