 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-2/12f/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

With

p = c lo g n / n

, the inequality

1 - p \leq e^{- p}

gives

E (N) = n (1 - p)^{n - 1} \leq n e^{- p (n - 1)} = n^{1 - c (n - 1) / n} ⟶ 0

when

c > 1

P (N > 0) \leq E (N) ⟶ 0.

P (N = 0) ⟶ 1 .

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