I’ve been working on this question, which I found on physics.SE. Unfortunately it was closed because it’s a homework question, but I’d like to get more of a hint than the original poster got.
I know that I’m supposed to use Bayes’ Theorem, but I don’t see how I’m supposed to use the fact that $\bar{N}$ is known. Every effort so far has yielded an unwieldy fraction that can’t be simplified.
Edit: Apparently I'm supposed to copy/paste the question. Here it is:
This is one of the exercises of Barnett's book on quantum information.
A particle counter records counts with an efficiency $\eta$. This means that each particle is detected with probability $\eta$ and missed with probability $1-\eta$. Let $N$ be the number of particles present and $n$ be the number of detected. Then: \begin{equation} P(n|N)=\frac{N!}{n!(N-n)!}\eta^n (1-\eta)^{N-n} \end{equation}
I know the mean number of particle present is:
\begin{equation} \bar{N}=\sum N P(N) \end{equation}
I want to calculate $P(N|n)$. I'm stuck here by a while, so I do not know how to proceed.
Edit 2: I'm adding a screenshot of the question in question:
You surely know that
$$ P(N|n) = \frac{P(n|N) P(N)}{P(n)}=\frac{P(n|N) P(N)}{\sum_N P(n|N) P(N)}$$
This tells you that you $P(n|N)$ is not enough, you need $P(N)$. To understand this should be your starting point.
Now, here, you only give us the mean of $N$ (actually you wrote the general formula of the expected value, but I guess you meant that you know $\bar{N}$). Still not enough.
Perhaps there is some implicit statistical model for $N$ that you've missed? Perhaps (just guessing) it follows a Poisson distribution? If so, you are done, because the Poisson distribution depends on a single parameter (which is the mean), hence you indeed can write $P(N)$.