Balls and Bins probability question?

Question

Given n balls and n bins, what is the expected number of bins with at least one ball given a random distribution?

Answer 1

OK, that's a tough one. The probablilty of a given ball NOT being in a given bin is (n-1)/n. The probability of ALL the balls not being in the bin is therefore ((n-1)/n)^n, or (n-1)^n/n^n

I think that ratio can also be used as the fraction of bins expected to be empty. Like, for instance, if the odds of a bin being empty were 1/8, then 1/8 of the bins should be empty. In which case, the fraction of non-empty bins should be 1-(n-1)^n/n^n. And the number of non-empty bins should be n*(1-(n-1)^n/n^n.

So I got n*(1-(n-1)^n/n^n.

Answer 2

34