There are $N$ elk, of which a simple random sample of size $n$ are captured and tagged ("simple random sample" means that all $\binom{N}{n}$ sets of $n$ elk are equally likely. The captured elk are returned to the population, and then a new sample is drawn, this time with size $m$. What is the probability that exactly $k$ of the $m$ elk in the new sample were previously tagged? (Assume that an elk that was captured before doesn't become more or less likely to be captured again.)

Can someone tell me why my approach is wrong (I compared it with the solution and it gives different numbers), here is what I tried:

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> After capturing and tagging the first batch, the elks now have a $\frac{n}{N}$ probability of being tagged.
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> Now, we recapture $m$ elk, thus the probability that $k$ out of $m$ are previously tagged is equal to the probability that $k$ are tagged multiplied by the probability that the remaining are not tagged.
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> That is,
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> $P = (\frac{n}{N})^k(\frac{N-n}{N})^{m-k}$
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**EDIT**: For those who are interested, to get the correct solution with my method it would be $\binom{m}{k}(\frac{n!(N-k)!}{(n-k)!N!})(\frac{(N-n)!(N-k-(m-k))!}{(N-n-(m-k))!(N-k)!})$. The reason it's so ugly is because I am sampling without replacement. As a result, the sample size decreases as we recapture the elk.

The first closed bracket is to represent that there are $\binom{m}{k}$ ways to pick the $k$ elks. The second closed bracket is the probability that $k$ are tagged. The third closed bracket is the probability that the remaining are not tagged.

For a better formula, I would look into hypergeometric probability as mentioned in Ian's answer.