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I am working through E.T. Jaynes Probability Theory book and am having difficulty with exercise 3.2: Suppose and urn contains $N = \sum N_i$ balls, $N_1$ of color 1, $N_2$ of color 2, ... $N_k$ of color $k$. We draw $m$ balls w/o replacement; what is the probability that we have at least one of each color?

Supposing $k = 5$ and $N_i = 10$, I reasoned the solution is $ {10 \choose 1}^5 / {50 \choose 5} $. How would I generalize this equation for $ m $ when $N_i$ and $k$ are not specified?

Thank you very much!

update: here is my solution to the generalized problem so far:
total number of possible draws: ${ N \choose m}$
combinations of $k$ colors: $\prod { N_i \choose 1} $
combinations of remaining draws: ${N - k \choose m -k } $
if the probability of m draws with at least k colors is:( (combinations satisfying at least one of k colors) $\times$ (combinations of rest of draws)) $\div$ (total number of draws) we get:
$ \frac{ \prod { N_i \choose 1}{N - k \choose m -k } }{{N \choose m}} = $ probability of drawing m balls with at least of each color k.

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