 Solution

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

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

Given

Z_{i} = n

, the probability of no friendship among people sharing day

i

2^{- n (n - 1) /2}

. The day counts are independent, so

P (no matching friend pair) = [e^{- 1} n = 0 \sum \infty \frac{2 ^{- n (n - 1) /2}}{n !}]^{365} = (\frac{2 + C}{e})^{365} .

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