Two different samples in Probability I was working on one exercise and i am not sure on the approach i should take there.
the exercise is the following:
A team of insect researchers was unsure whether the population size of a cetain severely endangered insect was $n_1=40$ or $n_2=50.$ They assess which of these two options is more likely, the following experiment was carried out: Initially, $4$ of these insects were captured, tagged, and released. Then, a few weeks later 6 of these insects were randomly captured, establised that 1 out of them had been previously tagged. Your task is to compute for $i=1$ and $2,$ the probability $\pi$ of the event "1 tagged species in a sample of 6" if the actual population is sizes $N_i.$ Which of $N_1$ and $N_2$ would in your opinion be a better option?
I seem to be stuck on the way i should solve this exercise.
 A: let event $E = $ "1 tagged species in a sample of 6"
$$P(\text{finding a insect tagged}\mid N_i) =\frac{4}{N_i}$$
$$P(E\mid N_i) = {6\choose1} \left(\frac{4}{N_i}\right)^1 \left(1-\frac{4}{N_i}\right)^5$$

I am basically  treating the event of sampling 6 insects as a bernoulli trial. The answer holds only if sampling is done with replacement as elicited by Michael Hardy in comments 
A: If you call the two numbers $n_1$ and $n_2$ you should not later call them $N_1$ and $N_2$ and act as if that were the same notation.
If the number of members of the species is $n$ and $4$ of them are tagged, then the probability that in a sample of $6$ there is exactly $1$ that is tagged is
$$
\frac{\dbinom 4 1 \dbinom{n-4}5}{\dbinom n 6} = \frac{4\cdot 6\cdot(n-6)(n-7)(n-8)}{n(n-1)(n-2)(n-3)}
$$
If you're using the method of maximum likelihood, you just figure out whether this is larger for $n=40$ or $n=50.$ However, if you have prior probabilities $p$ and $1-p$ assigned to those two possibilities, then you can proceed further with Bayes's formula.
