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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1/16/a/i/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 16 / a / i / Solution by

Codex  0 2026-10-18

Uniform sampling without replacement is equivalent to revealing a uniformly random permutation of seven labelled balls. Each draw position is equally likely to contain any label, so its colour has the original proportions. Thus

P (W_{1}) = \frac{3}{7}, P (B_{2}) = \frac{4}{7}, P (W_{3}) = \frac{3}{7} .

(1)

For example the law of total probability gives

P (B_{2}) = (4/7) (3/6) + (3/7) (4/6) = 4/7

. Equal marginal probabilities express exchangeability, rather than independence between successive draws.

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