There is a 20 card deck that contains 12 red cards, 4 black cards and 4 blue cards. You draw (without replacement) five cards. Let X be the number of red cards drawn and Y be the number of black cards drawn. Compute Cov (X,Y).
So I have the solution and everything makes sense but two things. I provided it below.
1. Are i and j are suppose to represent the number of color cards choosen for X and Y. For example i=j means there are the same number of red cards and black cards while i=1 means one red card choosen and j=2 means 2 black cards choosen.
2. I am confused by $E(X1Y1)=0$ since $P(X1Y1=0)=1$. I don't understand how the probability computes to this and how that makes the expectation 0.
$X_i$ is $1$ when the $i$-th card is red and $0$ otherwise, and $Y_i$ is $1$ when the $i$-th card is black and $0$ otherwise. The $i$-th card cannot be both red and black, so at least one of $X_i$ and $Y_i$ must be $0$, and it is therefore certain that $X_iY_i=0$; thus, $P(X_iY_i=0)=1$, and $$E(X_iY_i)=P(X_iY_i=0)\cdot0+P(X_iY_i=1)\cdot1=1\cdot0+0\cdot1=0\;.$$