There are n boxes numbered 1 to n and n balls numbered 1 to n. The balls are to be randomly placed in a box (not necessarily different boxes). What is the expected number of empty boxes?
I tried that:
$ X_i=1$ means Empty boxes
$ X_i=0$ otherwise
$E(\sum_{i=1}^{n} X_i)=\sum_{i=1}^{n}E(X_i)$.
P.S. We have that $E(X_i)=P(\{X_i=1\})=(({n-1}/n)^{n})$ for $i=1,\dots,n$
So the $E(\sum_{i=1}^{n} X_i)=n((({n-1})/n)^{n})$
I think I'm wrong but can't seem to get the correct idea. Any help would be appreciated.
For bin $k$, the probability no ball lands in this bin is $(1 - 1/n)^n$ by independence, since the probability the $i^{\text{th}}$ ball doesn't land there is $1 - 1/n$. Let $A_k$ denote the event that no ball lands in bin $k$. So we've computed that $\mathbb{P}(A_k) = (1-1/n)^n$.
The number of bins with no ball is $$\sum_{i=1}^n \mathbb{1}(A_k).$$ Thus the expected number of bins with no ball is $$\mathbb{E}\left[\sum_{i=1}^n \mathbb{1}(A_k)\right] = \sum_{i=1}^n \mathbb{E}\left[\mathbb{1}(A_k)\right] = \sum_{i=1}^n \mathbb{P}(A_k) = n(1-1/n)^n.$$