 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/ia/paper-2/11f/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

The statement is true. Put

C_{k} = ⋂_{r = 1}^{k} A_{r}

. The assumed

P (C_{n - 1}) > 0

ensures that every conditioning event

C_{k - 1}

2 \leq k \leq n

P (A_{k} ∣ C_{k - 1}) = \frac{P ( C _{k} )}{P ( C _{k - 1} )} .

Multiplying makes the fractions telescope, giving

\prod_{k = 2}^{n} P (A_{k} ∣ C_{k - 1}) = \frac{P ( C _{n} )}{P ( A _{1} )} = P (⋂_{k = 2}^{n} A_{k} ∣ A_{1}) .

This proves the chain rule for probabilities here without any assumption of independence. The final numerator

P (C_{n})

may be zero; all denominators remain positive.

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