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A committee of size $k$ is chosen from a group of $n$ women and $m$ men. All possible committees of size $k$ are equally likely. Let $X$ and $Y$ be the numbers of women and men on the committee, respectively.

(a) Find the joint PMF of $X$ and $Y$ . Be sure to specify the support.

(b) Find the marginal PMF of $X$ in two different ways: by doing a computation using the joint PMF, and using a story.

(c) Find the conditional PMF of $Y$ given that $X = x$.

I would have said that the joint PMF is hypergeometric $$P(X=x, Y=y) = \frac{\binom{n}{x}\binom{m}{y}}{\binom{n+m}{k}},$$ with $0\leq x \leq k$ and $y = k-x$, but then the other tasks make little sense. Do you have a hint for me on how to understand the problem?

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a) Correct.

b) Use $P\left(X=x\right)=P\left(X=x,Y=k-x\right)$ to find the PMF of $X$. A story? I don't know what they mean.

c) For $y\in\left\{ 0,\dots,k\right\} $ you have $P\left(Y=y\mid X=x\right)=\frac{P\left(X=x,Y=y\right)}{P\left(X=x\right)}$.

If $y=k-x$ then this results in $1$. If $y\neq k-x$ then it results in $0$. The distribution of $Y$ under condition $X=x$ has the characteristics of a distribution of a random variable constant at $k-x$.

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For part b:

b) $P(X=x)$= $\sum ^{k }_{y=0}$ $\binom{n}{x}$$\binom{m}{k-x}$/$\binom{n+m}{k}$= $\sum ^{k }_{y=0}$ $\binom{n}{k-y}$$\binom{m}{y}$/$\binom{n+m}{k}$= 1/$\binom{n+m}{k}$$\sum ^{k }_{y=0}$ $\binom{n}{k-y}$$\binom{m}{y}$,

where by Vandermonde

=$\binom{n+m}{k}$/$\binom{n+m}{k}$=$1$

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