 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/ii/solution

Use the law of total probability according to the first draw. Without replacement, the second-draw red probability is

1/6

after a red and

2/6

after a green. Thus

P (E_{2}) = \frac{2}{7} \frac{1}{6} + \frac{5}{7} \frac{2}{6} = \frac{2}{7} .

Equivalently, each counter has the same probability of occupying the second position as the first. Equal marginal probabilities do not make the two draw events independent.

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