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:

enter image description here

  • $\begingroup$ Please don't link a closed question. Copy the relevant (only math) parts here. $\endgroup$
    – leonbloy
    Commented Feb 3, 2019 at 2:28
  • $\begingroup$ @leonbloy I didn't know that was a rule, sorry. I've copied it now. $\endgroup$
    – NNN
    Commented Feb 3, 2019 at 4:43
  • $\begingroup$ "I want to calculate 𝑃(𝑁|𝑛)" To do so, you are missing some information, for example the (unconditional) distribution of N would do. $\endgroup$
    – Did
    Commented Feb 3, 2019 at 9:19

1 Answer 1


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 $P(N)$.

  • $\begingroup$ I've attached a screenshot of the exact question from the textbook. Unfortunately, doing this same calculation for a Poisson distribution was part (a) of the question. Do you have any other thoughts, given this? $\endgroup$
    – NNN
    Commented Feb 3, 2019 at 15:31
  • $\begingroup$ No, I don't see how one could do with the mean alone. $\endgroup$
    – leonbloy
    Commented Feb 4, 2019 at 15:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .