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I'm working through Stephen Barnett's book on quantum information and have come across the following question (1.5, for anyone keeping track at home)

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 detected. Show that

$$P(n|N) = \frac{N!}{(N-n)!n!}\eta^n(1-\eta)^{N-n}$$

a) Calculate P(N|n) for

$$P(N) = e^{-\bar{N}}\frac{\bar{N}^N}{N!}$$

b) Calculate P(N|n) for all P(N) equally probable

c) Calculate P(N|n) given only that the mean number of particles present is $\bar{N}$.

I've so far been able to get the first identity, plus part $(a)$.

My question is how I'm supposed to interpret $P(N)$, as written in question $(b)$. The use of capital $N$ as well as its use in part $(a)$ suggests that it could be "probability that $N$ particles are present". However if this the case then it can't be possible for $P(N)$ to be constant, since $N$ can be any natural number (infinitely many), and the sum of probabilities must be $1$. Maybe I'm just supposed to assume that the possible number of particles has an upper bound?

What am I missing here?

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