I'm having an issue with a probability problem concerning solutions.
Assume there is an "observational region" in a dilute solution with a volume $V$, and as solutes move across its boundary, the number $N$ of solute molecules inside the observation region fluctuates.
Divide $V$ into $M$ regions of volume $v$ each with $n$ particles. The solution is dilute enough that $n= 0$ or $1$ (there is no $v$ with more than one particle of solute), and each cell is occupied ($n = 1$) with probability $p = (ρ_0)v$.
If $W(N)$ is the number of configurations of the observation volume when $N$ solutes are present, what is the probability $P(N)$ of observing a given value of $N$, in terms of $p, W(N), M,$ and $N$.
I know the probability $P(n_1,n_2,\dots,n_M)$ of finding the system in a particular configuration in the observation volume is $p(N)=p^N (1−p)^{M−N},$ (Bernoulli Distribution), and since there are $N$ particles in $M$ spaces then the maximum number of configurations is $\frac{M!}{(N!(M-N)!)}$.
I'm not sure where to go from here.
If $p(N)$ is the probability of finding a given configuration of N particles, and $W(N)$ is the amount of configurations for a given $N$ number of particles, then the probability of observing $N$ particles total is the sum of the probability of finding each of the configurations that have $N$ particles. That is,
$$ P(N) = p(N) W(N) = p^N(1-p)^{M-N}W(N).$$
Every possible configuration has the same probability of occurring, so you just count how many of those meet your requisites (i.e., that they have $N$ particles).