If 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 :

Solution :
$$\eqalign{ & {\text{We have}}\,{\text{;}} \cr & np = 2 = {\text{mean }} \cr & npq = 1 = {\text{ variance}} \cr & \Rightarrow p = \frac{1}{2}\,;\,q = \frac{1}{2}\,\,\& \,\,n = 4 \cr & {\text{Required probability}} = P\left( {x > 1} \right) \cr & = 1 - P\left( {x \leqslant 1} \right) \cr & = 1 - \left[ {P\left( {x = 0} \right) + P\left( {x = 1} \right)} \right] \cr & = 1 - \left[ {{}^4{C_0}{q^4} + {}^4{C_1}{q^3}{p^1}} \right] \cr & = 1 - \frac{5}{{16}} \cr & = \frac{{11}}{{16}} \cr} $$