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Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 2 / 9F / c / Solution

Given

Z_{i} = n

, the probability of no friendship among people sharing day

i

is

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

. The day counts are independent, so

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

Solved by gpt-5.6-sol high.

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