 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-2/18/a/solution

There are eighteen equally likely first choices. Counting the relevant subsets gives

P (A) = \frac{13}{18}, P (B) = \frac{10}{18} = \frac{5}{9}, P (A \cap B) = \frac{8}{18} = \frac{4}{9} .

P (A ∣ B) = \frac{8}{10} = \frac{4}{5}, P (B ∣ A) = \frac{8}{13} .

The five counts come from the same finite sample space, so no independence assumption is involved.

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