 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2014/discrete-mathematics-continuous-mathematics-and-probability/7/a/ii/solution

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

Substitution proves Bayes' theorem:

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

The positive probabilities assumed in the question make both conditional expressions defined.

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