Finding the number of frogs in a pond A pond has an unknown number $N$ of frogs. To estimate $N$ Bob collects 40 frogs and marks
them. The next day, he collects 60 frogs, and finds that 13 of them are marked from the previous day.
Assume in both cases that the frogs are a random sample from the $N$.
Find the probability that the second sample has 13 marked frogs.
I can't think of what to do here, so a hint would be helpful. As in, what theorem, etc should I try to use.
 A: Just count combinations:
There are $N\choose 60$ samples of size 60 possible on day two, but only ${40\choose 13}\cdot{N-40\choose 20-13}$ of these samples have exactly 13 marked frogs.
Remark: Additionally to "Assume in both cases that the frogs are a random sample from the $N$" we need that the two samples are independent!
A: You have to calculate probability for each possible value of N knowing the circumstances, N could be 87 or higher
Probability for each N is $$p(N)=\frac{{40 \choose 13}{N-40 \choose  47}}{{N \choose 60}}$$
And the probability you're looking for is $\sum_{N=87}^{∞}p^2(N)$
A: You should use the hypergeometric distribution. Specifically, the number $X$ of marked frogs in the second sample is a random variable with hypergeometric distribution with paremeters


*

*$N=$ population size,

*$k=40$ successes, where success $=$ marked frog (there are $40$ from the previous day).

*$n=60$ sample size. 


Of course a reasonable assumption is that $N>n$. Hence, for any $$
$$P(X=x)=\frac{\dbinom{k}{x}\dbinom{N-k}{n-x}}{\dbinom{N}{n}}$$ For $x=13$ and for the above values for the parameters, you get that 
$$P(X=13)=\frac{\dbinom{40}{13}\dbinom{N-40}{60-13}}{\dbinom{N}{60}}$$ 
