 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-2/9f/i/solution

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

Put

Q = \sum_{j = 1}^{n} q_{j}

. Since the variables have continuous distributions, the minimum is unique almost surely. Integrating over its possible value gives

P (K = k, T \geq t) = \int_{t}^{\infty} q_{k} e^{- q_{k} s} j \neq = k \prod P (S_{j} > s) d s = \int_{t}^{\infty} q_{k} e^{- Q s} d s = \frac{q _{k}}{Q} e^{- Qt} .

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