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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 2 / 4F / b / Solution

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} .

Solved by gpt-5.6-sol high.

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