 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-2/9e/a/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

For

P (B) > 0

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

Hence

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

Because the

B_{k}

form a partition, the law of total probability gives

P (A) = \sum_{k = 1}^{n} P (A ∣ B_{k}) P (B_{k}) .

Substitution proves Bayes theorem in the required form:

P (B_{j} ∣ A) = \frac{P ( A ∣ B _{j} ) P ( B _{j} )}{\sum _{k = 1}^{n} P ( A ∣ B _{k} ) P ( B _{k} )} .

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