A bag has 10 red, 8 green, and 12 blue balls. I randomly grab 12 balls from the bag without replacement. Let $X$ be the number of red balls grabbed and $Y$ be the number of green balls grabbed. What is the covariance of $X$ and $Y$?
I began by creating a indicator random variable $I_i$ that is 1 if the $i$th ball grabbed is 1 and 0 otherwise as well as $J_i$, which is 1 if the $i$th ball grabbed is green and 0 otherwise. Then,
$ Covariance(I_1, J_1) = E(I_1 J_1) - E(I_1)E(J_1) = 0 - (\frac{10}{30})(\frac{8}{30}) \\ Covariance(I_1, J_2) = E(I_1 J_2) - E(I_1)E(J_2) = (\frac{10}{30})(\frac{8}{29}) - (\frac{10}{30})(\frac{8}{30} $
I got stuck here, and read the solutions, which read:
$ Covariance(X, Y) = 12 Covariance(I_1, J_1) + (12)(11) Covariance(I_1, J_2) = \frac{-96}{145} $
Where did this last equation come from? Why are we multiplying one of the covariances by 12 and the other one by 12 times 11?