 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-2/10f/d/solution

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

When

r

red marbles remain, the next draw reduces their number with probability

r / n

. The waiting time

W_{r}

{1, 2, \dots}

and therefore

G_{W_{r}} (s) = \frac{( r / n ) s}{1 - ( 1 - r / n ) s} = \frac{rs}{n - ( n - r ) s} .

The successive waiting times are independent, and

T = W_{n} + W_{n - 1} + \dots + W_{1} .

G_{T} (s) = r = 1 \prod n \frac{rs}{n - ( n - r ) s} .

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