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

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} .

ii

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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