An insured person is three times more likely to file exactly two claims than to file exactly three claims. Assume that the number X of claims filed has a Poisson distribution. how do you find E[X^2] ??
$\displaystyle \Pr(X = 2) = 3 \Pr(X = 3) \Rightarrow \frac{e^{-\lambda} \lambda^2}{2!} = \frac{3 e^{-\lambda} \lambda^3}{3!} \Rightarrow \lambda^2 = \lambda^3$. Solve for $\displaystyle \lambda$.
Now note that $\displaystyle E(X^2) - [E(X)]^2 = Var(X)$. And you should know that $\displaystyle E(X) = Var(X) = \lambda$ ....