Solution
= Solution
The empty-bin indicators are not independent. For distinct $i,k$,
$$
\mathbb P(Z_i=Z_k=1)=\left(1-\frac2n\right)^m,
$$
whereas $\mathbb P(Z_i=1)\mathbb P(Z_k=1)=(1-1/n)^{2m}$.
Each bin is empty precisely when all $m$ balls avoid it, so
$$
\mathbb EZ_i=\left(1-\frac1n\right)^m.
$$
The <linearity of expectation> does not require independence and gives
$$
\mathbb EZ=n\left(1-\frac1n\right)^m.
$$