 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-2/section-a/9/a/solution

There are

(2 7) = 21

equally likely unordered pairs, and

3 \cdot 4 = 12

contain one ball of each colour. Thus the hypergeometric distribution gives

P (different colours) = \frac{12}{21} = \frac{4}{7} .

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