= Solution
There are more bins than balls, so the <pigeonhole principle> forces at least one empty bin. In fact $E\geq m-n=(a-1)n$. Consequently
$$
\boxed{\mathbb P(E=0)=0.}
$$
For the expected fraction, substitute $m=an$ into part (b). The elementary exponential limit, obtained by taking logarithms, gives
$$
\boxed{\frac{\mathbb E[E]}m=\left(1-\frac1{an}\right)^n\longrightarrow e^{-1/a}.}
$$
This is the empty-bin case of the <Poisson limit for occupancy fractions>, with limiting mean occupancy $1/a$.
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