# Distribution of number of special elements chosen: m choices of n items with k special items

Suppose I a set with $n$ items, $k$ of which have a certain property($k\leq n$), and I choose $m$ items randomly from that set($m\leq n$), what is the distribution of the number of chosen items having that property? An item can only be chosen once.

I expect an answer in the form $\left( \begin{array}{c} f(k,n,m) \\ g(k,n,m) \end{array} \right)$, but I don't know $f(k,n,m)$ and $g(k,n,m)$.

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Begin by counting all samples of $m$ elements from the population of $n$; this can be done $n \choose m$ ways. Suppose exactly $l$ of the $k$ special items are chosen; let us enumerate the possibilities. There are $k\choose l$ ways to choose the special items and then $n-k\choose m -l$ ways to choose the non-special ones. Now divide; here we assume all samples are equiprobable. We get
$$P(\hbox{exactly l special elements}) = {{k \choose l}{n - k\choose m -l}\over{n\choose m}}\qquad 1\le k \le m$$ This is called a hypergeometric distribution.