 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-2/9f/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Urn

i

has

2^{i}

balls, so a draw from it is white with probability

2^{- i}

. Given that the first ball is white, Bayes theorem gives the posterior probability

P (I = i ∣ W) = \frac{2 ^{- i}}{\sum _{j = 1}^{n} 2 ^{- j}} .

After replacement, the second draw is conditionally independent of the first given the urn. Hence

p (n) = \frac{\sum _{i = 1}^{n} 4 ^{- i}}{\sum _{i = 1}^{n} 2 ^{- i}} = \frac{\frac{1}{3} ( 1 - 4 ^{- n} )}{1 - 2 ^{- n}} = \frac{1}{3} (1 + 2^{- n}) .

Consequently

p (n) \to 1/3

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