 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-1/16/c/i/solution

The event of two blue draws splits into disjoint replacement and no-replacement cases. Their probabilities from the sample space are

4 r /25

and

(1 - r) /10

. Hence

P (B_{1} \cap B_{2}) = \frac{4 r}{25} + \frac{1 - r}{10} = \frac{5 + 3 r}{50} .

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