Covariance problem involving and urn and some balls Okay so here is the question - Suppose we draw five balls from an urn without replacement. The urn has $5$ red, $6$ black, and $5$ green balls. If $X$ gives the number of green balls drawn and and $Y$ the number of red balls drawn, find $Cov(X,Y)$. 
Here's my thought process so far $Cov(X,Y) = \mathbb{E}(XY) - \mathbb{E}(X)\mathbb{E}(Y)$ so then I set it up into $Y_i$ {$1$ if $i^{th}$ draw is red, $0$ ELSE and then $X_i$ {$1$ if i$^{th}$ draw is green, $0$ ELSE}.
From here I changed it to $Cov(Y_i,X_i) = \mathbb{E}(Y_i X_i)-\mathbb{E}(Y_i)\mathbb{E}(X_i)$. This is as far as I can get and I don't even know if I'm on the right track here. Can anyone help me out?
 A: You are indeed on the right track, except that you have to consider all $X_i$ and $Y_j$, not just $X_i$ and $Y_i$.  Next step: $\mathbb E(X_i Y_i) = 0$ because the $i$'th draw can't be both red and green.
What are $\mathbb E(X_i)$ and $\mathbb E(Y_i)$?
Another way of doing it is to consider the indicators for individual green and red balls, rather than draws.
A: I think you are on the right track.  As a general rule, $$Cov(\sum_{i=1}^nX_i,\sum_{j=1}^nY_j)=\sum_{i=1}^n\sum_{j=1}^nCov(X_i,Y_j)$$
When $i=j, Cov(X_i,Y_j)=E(X_iY_j)-E(X_i)*E(Y_j)=0-({5\over 16}*{5\over 16})=-{25\over 256}$
Note that in this case $E(X_iY_j)=0$ since the same ball can't be red and green.
When $i\ne j, Cov(X_i,Y_j)=E(X_iY_j)-E(X_i)*E(Y_j)=({5\over 16}*{5\over 15})-({5\over 16}*{5\over 16})={5\over 768}$
Note that in this case $E(X_iY_j)={5\over 16}*{5\over 15}$ since the chance of any given ball being red is $5\over 16$ and the odds of a different ball being green is then $5\over 15$
We must now add all the covariances together and since $i=j$ five times and $i\ne j$ 20 times, we have $$Cov(X,Y)=5*-{25\over 256}+20*{5\over 768}=-{275\over 768}$$
