 Solution

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/iii/solution

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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