 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2001/paper-2/11/b/v/solution

P (E_{1} ∣ E_{2}) = \frac{P ( E _{1} \cap E _{2} )}{P ( E _{2} )} = \frac{1/21}{2/7} = \frac{1}{6} .

Conditioning on a red in the second position leaves one red among the six counters that could occupy the first position, giving the same result directly.

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