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Suppose that $10$ balls are put into $5$ boxes, with each ball independently being put in box $i$ with probability $ p_i, \sum_{i=1}^{5} p_i = 1 $

A) Find the expected number of boxes that do not have any balls.

Attempt: Let $X$ denote the number of boxes without balls. This means $ EX = \sum_{x=0}^{4} x P(X=x) $ Since we know each ball must go into at least one box, we cannot have 5 empty boxes, so that is why I sum to 4. I then said $P(X=j) = P(j{}\,\text{boxes with no balls})= {5 \choose j}(1-p_i)^{10}$ So $EX = 0 + \sum_{j=1}^{4} j{5 \choose j}(1-p_i)^{10}$ Is it ok?

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    No. It makes no sense since it contains a free index $i$.2012-12-06

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The probability that box $i$ remains empty is $(1-p_i)^{10}$. By linearity of expectation, the expected number of boxes that remain empty is $\sum_i(1-p_i)^{10}$.

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    @CAF: I'm sorry, I don't know how to make it any clearer. A non-empty box is $0$ empty boxes and an empty box is $1$ empty box. This is not representing, just a trivial form of counting. Try thinking less complicated, and it will suddenly become very obvious.2012-12-12