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 one is equal to

\(\begin{aligned} & \text { Mean }=2 \text { and variance }=1 \\ & \Rightarrow \mathrm{np}=2 \text { and } \mathrm{npq}=1 \\ \therefore \quad & \frac{\mathrm{npq}}{\mathrm{np}}=\frac{1}{2} \Rightarrow \mathrm{q}=\frac{1}{2} \\ & \Rightarrow \mathrm{p}=1-\frac{1}{2}=\frac{1}{2} \\ & n p=2 \\ & \Rightarrow \mathrm{n}\left(\frac{1}{2}\right)=2 \Rightarrow \mathrm{n}=4\end{aligned}\)
\(\begin{array}{ll}\therefore & P(X\gt1) \\ & =P(X=2)+P(X=3)+P(X=4)\end{array}\)
\(={ }^4 \mathrm{C}_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^2+{ }^4 \mathrm{C}_3\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)+{ }^4 \mathrm{C}_4\left(\frac{1}{2}\right)^4\left(\frac{1}{2}\right)^0\)
\(\begin{aligned} & =6\left(\frac{1}{2}\right)^4+4\left(\frac{1}{2}\right)^4+1\left(\frac{1}{2}\right)^4 \\ & =\frac{1}{2^4}(6+4+1)=\frac{11}{16}\end{aligned}\)