 Solution

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

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

Setting

t = 0

in part (i) gives

P (K = k) = \frac{q _{k}}{Q} .

Also

P (T \geq t) = \prod_{j = 1}^{n} e^{- q_{j} t} = e^{- Qt},

T \sim Exp (Q) .

Part (i) now factors as

P (K = k, T \geq t) = P (K = k) P (T \geq t),

which proves that

K

and

T

are independent.

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