 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/iii/solution

After a red first draw, one red remains among six counters. The multiplication rule for conditional probability gives

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

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