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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:

After capturing and tagging the first batch, the elks now have a $\frac{n}{N}$ probability of being tagged.

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.

That is,

$P = (\frac{n}{N})^k(\frac{N-n}{N})^{m-k}$

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.

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  • $\begingroup$ The formula in your edit is exactly $\binom{n}{k} \binom{N-n}{m-k} / \binom{N}{m}$. There are $\binom{N}{m}$ ways to choose $m$ elk out of $N$, so that's the denominator. Now, we want exactly $k$ elk from the previously-tagged set, and there are $\binom{n}{k}$ ways to choose these elk. Then we also need an additional $(m-k)$ elk from the un-tagged set, and there are $\binom{N-n}{m-k}$ ways to make this happen. Note: the FIRST sample didn't have to be random at all; the same analysis works for ANY previously-chosen set of $n$ elk. Only the second sample has to be random. $\endgroup$ – mathmandan May 20 '16 at 5:04
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I think your logic would actually imply that the number of tagged elk in the new sample is a binomial random variable with success probability $n/N$. This would need a factor of ${m \choose k}$ in addition to what you wrote. However, that can't possibly be right because it is impossible to have $k>n$, and yet this method doesn't see anything change as $k$ passes through $n$.

The problem is that you need to sample without replacement in the second sample (a trapped elk is temporarily out of the population). This means that you have a population of $N$ things, $n$ of which count as successes and $N-n$ of which count as failures, and you want to take a sample of size $m$ without replacement and consider the probability of exactly $k$ successes in that sample. This is described by the hypergeometric distribution, though if you didn't know that then you could derive the hypergeometric distribution using counting principles.

If $m \ll n$ and $n$ is not too close to $0$ or to $N$, then your answer would be approximately binomial as you wrote (but with the factor of ${m \choose k}$ in front), because as you proceeded to sample the fraction of tagged elk in the population would not change appreciably.

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  • $\begingroup$ Wow that makes so much more sense, thank you so much Ian. For those who are interested, to get the correct solution with my method it would be $\binom{m}{2}(\frac{n!(N-k)!}{(n-k)!N!})(\frac{(N-n)!(N-k-(m-k))!}{(N-n-(m-k))!(N-k)!})$. Super ugly, but it works haha :). Also updated my question with an edit that explains how it works. $\endgroup$ – Zoom Bee May 19 '16 at 23:55
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There are $m \choose k$ ways to pick a set of $k$ elks from the sample of size $m$ that could have been previously tagged. Thus, the answer should be ${m \choose k}\left(\frac n N\right)^k\left(\frac {N-n}{N}\right)^{m-k}$.

These kinds of problems are binomial probability problems and in these kinds of problems, it's important to remember the factor of $m \choose k$ because otherwise, you will not get the right probability.

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