 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ia/paper-2/9f/a/solution

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

P (B_{i} ∣ A) = \frac{P ( A \cap B _{i} )}{P ( A )} = \frac{P ( A ∣ B _{i} ) P ( B _{i} )}{P ( A )} .

Since the

B_{j}

form a disjoint partition of the sample space, the law of total probability gives

P (A) = \sum_{j} P (A ∣ B_{j}) P (B_{j})

. Substitution proves Bayes' theorem in the required form:

P (B_{i} ∣ A) = \frac{P ( A ∣ B _{i} ) P ( B _{i} )}{\sum _{j} P ( A ∣ B _{j} ) P ( B _{j} )} .

The hypotheses make the denominator positive and each conditioning probability well defined.

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