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
$$
\mathbb EZ=n\left(1-\frac1n\right)^m.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 208 / 3 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-208/3/a/solution