Suppose m balls randomly and independently put into n boxes. A box can hold more than 1 ball. Then what is the expected number of empty boxes? What is the expected number of balls that in a box with at lease one other ball?
For the first question, I don't understand why $P(X_i) = (\frac {n-1}{n})^m$ (this is the solution), where $X_i$ denotes event in which the $i^{th}$ box is empty. I think $P(X_i) = \frac {n-1}{n}$, since one ball has $n$ boxes to put in except the $i^{th}$. So where is the power of $m$ comes from?
I don't even have a clue how to solve the second question.
Best Answer
As JMoravitz has already commented, the probability $\frac{n-1}n$ of a single ball not being placed in a particular box is taken to the $m$-th power because we need the probability of all $m$ balls not being placed in that particular box, and by the multiplication principle we need to multiply the $m$ individual probabilities.
For the second question, for a particular ball the probability that none of the other $m-1$ balls are placed in its box is $\left(\frac{n-1}n\right)^{m-1}$, so the expected number of balls in a box with at least one other ball is
$$ m\left(1-\left(\frac{n-1}n\right)^{m-1}\right)\;. $$