The prior probabilities of choosing the two urns are equal. Their white-ball probabilities are

P (W ∣ I) = \frac{45}{75} = \frac{3}{5}, P (W ∣ II) = \frac{12}{48} = \frac{1}{4} .

P (I ∣ W) = \frac{3/5}{3/5 + 1/4} = \frac{12}{17} .

ii

Solution

The two possible ways to observe one ball of each colour have probabilities

P (W_{I}, B_{II}) = \frac{3}{5} \frac{3}{4} = \frac{9}{20}, P (B_{I}, W_{II}) = \frac{2}{5} \frac{1}{4} = \frac{2}{20} .

Conditioning on their union therefore gives

P (W_{I}, B_{II} ∣ one white and one black) = \frac{9}{11} .

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