Computing the variance of hypergeometric distribution using indicator functions

$newcommand{var}{operatorname{var}}newcommand{cov}{operatorname{cov}}$Just a small note to Michael's answer. The number of $2 cov(I_{A_{1}}, I_{A_{2}})$ terms is $nchoose 2$. Thus, the variance becomes:

begin{align} var(I_{A_1}+cdots+I_{A_n}) & = var(I_{A_1})+cdots+var(I_{A_n}) + underbrace{2cov(I_{A_1},I_{A_2})+cdotsquad{}}_{{nchoose 2}text{ terms}} \[10pt] & = nvar(I_{A_1}) + {nchoose 2} 2 cov(I_{A_1},I_{A_2}). end{align}

(I wrote it as a separate answer, because it was rejected as an edit, and don't have enough reputation to comment.)

Outline: I will change notation, to have fewer subscripts. Let $Y_i=1$ if the $i$-th ball is red, and let $Y_i=0$ otherwise.

We are picking $n$ balls. I will assume that (unlike in the problem as stated) $n$ is not necessarily the total number of balls, since that would make the problem trivial.

Then $E(X)=E(Y_1)+cdots+E(Y_n)$. Note that $Pr(Y_i=1)=frac{r}{r+b}$. For if the balls have ID numbers (if you like, in invisible ink) then all sequences of balls are equally likely.

For the variance, as you know, it is enough to compute $E(X^2)$. Expand $(Y_1+cdots+Y_n)^2$ and take the expectation, using the linearity of expectation.

We have terms $Y_i^2$ whose expectation is easy, since $Y_i^2=Y_i$. So we need the expectations of the "mixed" products $Y_iY_j$. We need to find the probability that the $i$-th ball and the $j$-th ball are red. This is the probability that the $i$-th is red times the probability that the $j$-th is red given that the $i$-th is.

Thus $E(Y_iY_j)=frac{r}{r+b}cdotfrac{r-1}{r+b-1}$.

Now it s a matter of putting the pieces together.