 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-2/4f/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

At station

j

, the probability of stopping is

\frac{2}{n} - \frac{1}{n} (1 - \frac{1}{n})^{j - 1},

because the lower rate applies exactly when no earlier stop occurred. By linearity of expectation,

E X = 2 - \frac{1}{n} \sum_{j = 1}^{n} (1 - \frac{1}{n})^{j - 1} = 1 + (1 - \frac{1}{n})^{n} ⟶ 1 + e^{- 1} .

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