 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2018/discrete-mathematics-continuous-mathematics-and-probability/8/a/ii/solution

The all-red event

B

is a subset of the same-colour event

A

(1/2)^{3} = 1/8

, while the three disjoint monochromatic cases give

P (A) = (\frac{1}{2})^{3} + (\frac{1}{3})^{3} + (\frac{1}{6})^{3} = \frac{1}{6} .

P (B ∣ A) = \frac{P ( B )}{P ( A )} = \frac{3}{4} .

The monochromatic outcomes are not equally likely, so counting colours alone would give the wrong conditional answer.

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