 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1/16/a/ii/solution

The first black draw has probability

4/7

. Given that event, three of the remaining six balls are black. The multiplication rule therefore gives

P (B_{1} \cap B_{2}) = \frac{4}{7} \frac{3}{6} = \frac{2}{7} .

Using

(4/7)^{2}

would incorrectly treat the draws as independent.

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