 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/ia/paper-2/10f/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

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

P (E = 0) = 0.

For the expected fraction, substitute

m = an

into part (b). The elementary exponential limit, obtained by taking logarithms, gives

\frac{E [ E ]}{m} = (1 - \frac{1}{an})^{n} ⟶ 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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