 Solution

For each of the $b$ balls, let $I_i$ be the indicator random variable that ball $i$ survives the round of uniform balls-in-bins allocation. Condition on its chosen bin. Each of the other $b-1$ balls avoids that bin with probability $1-1/n$, independently, so the probability that ball $i$ is served is $(1-1/n{)}^{b-1}$. Hence, for $b\ge 2$,We used linearity of expectation to sum the indicator random variables; their mutual independence is unnecessary. For $b=0$ or $b=1$, the next count is zero. This states the boundary cases without an ambiguous $0^0$ when $n=1$.