 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-2/17/c/i/solution

P (A ∣ B) P (B) = P (A \cap B)

. Applying it once more gives

P (A ∣ B) P (B) P (C ∣ A \cap B) = P (A \cap B \cap C) .

The conditionals in the original expression require their conditioning events to have positive probability.

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