There are N fish in a pond. You use a net to catch them, and each fish is equally likely to be caught. One day, you catch 40 fish, tag each of them, and then throw them back. The next day, you catch 30 fish in the net and note that 6 have tags.
What is your best guess of the value of N?
I think this relates to the capture - recapture problem where N= total population; n=first sample, k=tagged items, m=second sample.
$\frac{\binom{n}{k}\binom{N-n}{m-k}}{\binom{N}{m}}$
but I honestly do not know how to proceed.
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$\begingroup$Keeping it simple:
on Day 1 proportion of fish in pond that are tagged $=\frac{40}{N}$
on Day 2 proportion of fish in sample that are tagged $=\frac{6}{30}$
$\endgroup$ 1 $\begingroup$so solve $\frac{40}{N}$ = $\frac{6}{30}$ to get N = 200
As you said, we need to use the capture-recapture formula, but this does not involve combinatorics. Here is the formula below: $$N=\frac{1st\:sample\:\times\:2nd\:sample}{2nd\:sample\:previously\:marked}$$ So in this case, $N =(40×30)/6=1200/6=200 $
$\endgroup$ 3 $\begingroup$Define a function $f:\mathbb{Z}_{\geq 64}\rightarrow [0,1]$ by $f(n)=\frac{{40 \choose 6}{n-40 \choose 24}}{{n \choose 30}}$. Then $f(n)=P(X=6|N=n)$ where $X$ is the number of tagged fish in your second sample of $30$ fish. I would use the value of $n$ that maximizes $f(n)$ as my estimate for $N$.
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