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I think the easiest approach is to determine the distribution of the number of urns containing green balls. Conditioned on these cases, you can compute the probability that the black balls are in the other urns.


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But I dont really know how to come up with the distribution for the 6 balls.

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You don't need this distribution to compute the answer. You can use the distribution of the number of urns containing at least one of the 4 green balls, which is simple enough to compute in an ad hoc fashion. For example, you can compute the probability that precisely two urns contain a green ball, then compute the probability that the 6 black balls are in the other 2 urns. It's easier to compute 1, 2, and 4, and then you can subtract these probabilities to get 3.

If you want a more general method to determine the distribution, you can imagine placing the balls of one color in the urns one at a time while keeping track of the number of urns containing at least one. If you have U urns and N are occupied, the probability of moving to the state of N+1 occupied is (U-N)/U, and the probability of staying at N occupied is N/U. You can encode these transitions in a transfer matrix, and adding B balls corresponds to applying the Bth power of the transfer matrix. It's also easily computable by hand when B is small.

These seem like combinatorics problems, not probability problems.