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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/2020/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 / 2020 / Paper 1 / 16 / a / i / Solution by

Codex  0 2026-10-18

Random sampling without replacement assigns each of the ten distinguishable pieces to each lunch with the same probability. Five pieces of each type therefore give

P (A_{1}) = \frac{1}{2}, P (B_{2}) = \frac{1}{2} .

(1)

One can check the second probability by conditioning on the first lunch:

P (B_{2}) = P (B_{2} ∣ A_{1}) P (A_{1}) + P (B_{2} ∣ B_{1}) P (B_{1}) = \frac{5}{9} \frac{1}{2} + \frac{4}{9} \frac{1}{2} = \frac{1}{2} .

(2)

The identical marginal probabilities do not imply independence of the draws.

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