Say we have N balls and K bins. Let's call Y - The number of balls in the last bin. What is E(Y) ?
I don't know that way to get E(Y), I think there is a way of finding it without using indicators.
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$$Y\sim B(N,1/K)$$ – Kenny Lau May 07 '16 at 17:03
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The question is just full on incomplete! (I guess everybody gets what you mean, but still it just looks hasty and incomplete) – Patrick Abraham May 07 '16 at 17:09
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Possible duplicate of Expectaion of a balls and bins question – callculus42 May 07 '16 at 17:22
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Assume that each ball is different (if they are the same, make them different).
The probability that one ball goes to the last bin is $\frac1K$.
Therefore, $Y\sim B\left(N,\frac1K\right)$.
Also, the expected value of a binomial distribution with parameters $n$, $p$ is $np$.
Therefore, the expected value of $Y$ is $\frac NK$.
