 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-2/17/a/ii/solution

P (X ∣ R) = \frac{P ( R ∣ X ) P ( X )}{P ( R )} = \frac{( 20/50 ) ( 1/2 )}{16/35} = \frac{7}{16} .

The denominator sums red outcomes from both possible jar choices, as calculated in (i).

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