 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/ii/solution

P (R ∣ B_{1} \cap B_{2}) = \frac{P ( B _{1} \cap R \cap B _{2} )}{P ( B _{1} \cap B _{2} )} = \frac{4 r /25}{( 5 + 3 r ) /50} = \frac{8 r}{5 + 3 r} .

Its denominator is positive throughout the specified range. Observing two blue balls raises the probability of replacement for

0 < r < 1

, because replacement makes a repeated colour more likely.

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