Poisson random variable- variance Let us model the number of winter storms in a given year as a Poisson random variable. Suppose that in a good year the average number of storms is 3, and in a bad year the average is 5. If the next year will be good with probability 40% and bad with probability 60%, find the expected number of winter storms next year.
E[X] = 5*.6 + 3*.4 = 4.2 <-- correct answer
Next, find the variance of the number of winter storms next year.
I know that Var[X] = E[X^2] - E[X}^2 so when I calculated this I got:
E[X^2] = 5^2(.6) + 3^2(.4) = 18.6 and E[X]^2 = 4.2^2 = 17.64 so Var[X] = 18.6 -17.64 = .96.
However, this is not the correct answer. Any ideas of where I am going wrong?
 A: Let me clarify what you did wrong. (1) undefined variables and (2) calculating squared expectation incorrectly
If we proceed as you did, but set up the variables and notation first, we go something like this:
Let $B$ be $0$ in a good year and $1$ in a bad year. 
($B\sim\text{Bernoulli}(0.6)$)
Let $X$ be the number of storms next year.
Let $S_0$ be the number of storms in a good year. Let $S_1$ be the number of storms in a bad year.
$S_0=(X|B=0)\sim\text{Pois}(3)$
$S_1=(X|B=1)\sim\text{Pois}(5)$
Then 
$$E(X) \,\,= E(S_0)\times (1-p_B)+E(S_1)\times p_B$$
$$\quad = E(S_0)\times 0.4+E(S_1)\times 0.6$$
as you calculated before (this is just the law of total expectation)
$$E(X^2) \, = E(S_0^2)\times (1-p_B)
+E(S_1^2)\times p_B$$
$$\quad = E(S_0^2)\times 0.4+E(S_1^2)\times 0.6$$
(this is the same formula you used for expectation, but now in terms of random variables that are the squares of the ones you had before)
Lack of notation aside, you actually are okay up to here. 
The mistake is that you (implicitly) then used the variances in place of $E(S_i^2)$, which is too small. You need to add another term!
I think the problem was your lack of defined variables obscured your mistake, making it impossible for you to find.
If you do that bit right, you'll get an $E(X^2)$ which is between 22 and 23 (and is getting up toward 23 as I mentioned before).
That approach works. But to my mind the law of total variance is easier. If you've never encountered it, do the $E(X^2)$ approach - carefully.
