foundations in probability

Let's start by finding the covariance between X and Y. Refer to table1 I've uploaded. That is the joint distribution of X and Y. Then look at table 2. It's the distribution of X times Y. Let's use that to calculate the expected value of X*Y, which turns out to be 9/8.

X ~ binomial(n = 3, p = .5), therefore E(X) = n*p = 3/2, and VAR(X) = n*p*(1-p) = 3/4

Y ~ binomial (n = 2, p = .5), therefore E(Y) = n*p = 1, and VAR(Y) = n*p*(1-p) = 1/2

COV(X,Y) = E(XY) - E(X)*E(Y) = 9/8 - 3/4 * 1 = 3/8

finally, COR(X,Y) = COV(X,Y)/sqrt(VAR(X) * VAR(Y)) = (3/8)/sqrt(3/4 * 1/2) = sqrt(6)/4 $\approx$ 0.6124.