 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-2/9f/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Conditional on

N = n

, there are

(2 n)

pairs, and each is both friendly and birthday-matched with probability

1/ (2 \cdot 365)

. Since

E [N (N - 1)] = 36 5^{2}

for a Poisson variable,

E [matching friend pairs] = \frac{36 5 ^{2}}{2} \frac{1}{730} = \frac{365}{4} .

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