 Solution

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

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

By Poisson thinning, allocating a Poisson

(365)

population independently and uniformly among

365

days gives independent Poisson

(1)

counts. Hence

P (Z_{1} = z_{1}, \dots, Z_{365} = z_{365}) = e^{- 365} i = 1 \prod 365 \frac{1}{z _{i} !}, z_{i} \in Z_{\geq 0} .

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