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 / iii

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- Solution iii

Solution

Any two distinct draw positions have the same pair law by exchangeability. The marginal and pair probabilities in sampling without replacement therefore give

P (B_{2} \cap B_{3}) = \frac{4 \cdot 3}{7 \cdot 6} = \frac{2}{7} .

Divide this joint probability by the relevant marginal probability to get the two conditional probabilities:

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

Conditioning on the third draw can change the probability of the second draw; conditional probability does not require the conditioning event to occur earlier in time.

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