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I have $n$ bins such that each bin may contain at most $2$ balls. we start inserting $k$ balls into the bins, by choosing randomly for each ball the bin to which it will be inserted. what is the probability that some designated bin of a ball will be full (i.e. will contain $2$ balls already)?

My idea was: the probability of a bin to be full (for some ball) is the chance that some other ball was already inserted to that bin (i.e. $\frac{k}{n}$) multiplicated by the chance that one more ball was inserted to the same bin (i.e. $\frac{k-1}{n}$) as we have $k-1$ more balls to distribte between the bins. However, something seems to be missing, I think this is not the correct answer. can anyone recognize the problem?

Edit: The bins already contain k balls. The question is what is the probability that the bin that was randomly chosen for some new ball (equal probability for all n bins) was full (already contained 2 balls)?

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  • $\begingroup$ I don't understand how the balls are placed in the bins. $\endgroup$ – Jorge Fernández Hidalgo Mar 27 '16 at 17:14
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Your probability only takes into account a certain arbitrary ball. You need to consider that this could happen for every single ball.

Hint: try to put one ball at a time, you will get a sum. What is the closed form of the sum?

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