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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?

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Let Abe pick his $50$ friends in Springfield first. Then let Zach pick $50$ friends among the $1000$ people in Springfield, and let $X$ be the number of these that are also Abe's friends. Since Zach picks his friends without replacement, $X$ will have the hypergeometric distribution. Hence $$P(X=k)={{50\choose k}{950\choose 50-k}\over{1000\choose 50}}, \qquad k=0,\dots,50$$

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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 $\binom{999}{49}$.

$$E[X]=\frac{50^2}{1000}=2.5.$$

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