 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-2/9e/a/ii/solution

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

Urn

r

has

n - r

blue balls among

n - 1

balls. Averaging over the uniformly chosen urn,

P (first blue) = \frac{1}{n} \sum_{r = 1}^{n} \frac{n - r}{n - 1} = \frac{1}{2} .

The probability that both removed balls are blue is

P (B_{1} \cap B_{2}) = \frac{1}{n} r = 1 \sum n \frac{( n - r ) ( n - r - 1 )}{( n - 1 ) ( n - 2 )} = \frac{1}{3} .

Therefore

P (B_{2} ∣ B_{1}) = \frac{P ( B _{1} \cap B _{2} )}{P ( B _{1} )} = \frac{2}{3} .

The first blue draw makes urns with more blue balls more likely, which explains why the conditional probability exceeds one half.

 New to topics? Read the docs here!