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So suppose we have N balls, and M bins with size $S_i$ for $1\leq i \leq M $.

I know that if all bins have the same size, then the probability of finding a given number of $n_i$ balls in bin $B_i$ for $1\leq i\leq M$ is

$P = \frac{N!}{n_1! n_2! ...n_m!}$.

My question is how would this formulation if we have bins of different sizes. My gut says it would be $P = \frac{N!}{\big(\frac{n_1}{S_1}\big)! \big(\frac{n_2}{S_2}\big)! ...\big(\frac{n_m}{S_m}\big)!}$.

If someone could help me on this, I would greatly appreciate it.

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  • $\begingroup$ Your probability $P$ (the first time written) exceeds $1$ $\endgroup$ – drhab Oct 6 '18 at 16:05
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If all spots have equal probability to be the spot where a ball is located then you can restate it like this:

If there are $S:=\sum_{i=1}^m S_i$ balls in an urn and for $i=1,\dots,m$ there are $S_i$ balls that have color $i$, then what is the probability of (again for $i=1,\dots,n$) drawing $n_i$ balls of color $i$ if in total $n=\sum_{i=1}^Mn_i$ balls are drawn?

The answer is: $$\frac{\binom{S_1}{n_1}\times\cdots\times\binom{S_m}{n_m}}{\binom{S}n}$$An application of multivariate hypergeometric distribution.

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