 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2005/paper-1/3/b/ii/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

N - 1

other guests avoids the host's date with probability

364/365

P (at least one matches the host) = 1 - (\frac{364}{365})^{N - 1} .

For this to exceed

1/2

, the negative logarithm of

364/365

reverses the inequality on division, giving

N - 1 > ln (1/2) / ln (364/365) ≃ 252.652

. Thus the smallest total attendance is $N = 254$ , including the host.

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