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

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- Solution 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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