Maximum Variance of a Distribution

Question

Let the random variable $X$ have the distribution $\mathbb{P}(X=0)=\mathbb{P}(X=2)$, $\mathbb{P}(X=1)=1-2p$ for $0\leq p \leq 1/2$. For what $p$ is the $\mathrm{Var}(X)$ maximum ?

Answer 1

By definition of expectation, $E(X) = 0*P(X=0)+1P(X=1)+2P(X=2)$. Note that $P(X=0) + P(X=2) = 1-P(X=1)=2p$. Thus, $P(X=0)=P(X=1)=p$ and $E(X) = 0*p + 1*(1-2p) + 2p = 1$. Furthermore, $E(X^2) = 1*(1-2p) + 4*p=1+2p$. The variance, $V(X)=E(X^2)-(E(X))^2$ is maximal for $p=1/2$.