Find the probability mass function. Abe and Zach live in Springfield. Suppose Abe's friends and Zach's friends are each a random sample of 50 out of the 1000 people who live in Springfield.
Find the probability mass function of them having $X$ mutual friends.
I figured the expected value is $(1000)(\frac{50}{1000})^2 = \frac{5}{2}$ since each person in Springfield has a $(\frac{50}{1000})^2$ chance of being friends with both Abe and Zach. However, how do I generalize this expected value idea to create a probability mass function that returns the probability of having $X$ mutual friends?

The expected value is right, since:
$$E[X]=\frac{1}{\binom{1000}{50}}\sum_{k=1}^{50}k\binom{50}{k}\binom{950}{50-k}=\frac{50}{\binom{1000}{50}}\sum_{k=0}^{49}\binom{49}{k}\binom{950}{49-k}$$
$$\frac{\binom{1000}{50}}{50}\,E[X]=[x^{49}]\left((1+x)^{49} x^{49} (1+x^{-1})^{950}\right)=[x^{49}]\left((1+x)^{999}x^{-901}\right)=[x^{950}](x+1)^{999}=\binom{999}{950}=\binom{999}{49},$$
or, more simply, $\frac{\binom{1000}{50}}{50}\,E[X]$ is the number of ways to choose 49 stones between 49 black stones and 950 withe stones, so it is clearly $\binom{999}{49}$.
$$E[X]=\frac{50^2}{1000}=2.5.$$
