Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 2 / 11 / a

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- Solution a

Solution

With replacement, successive draws are independent, with red probability

2/7

and green probability

5/7

. The first red on draw

n

requires

n - 1

greens followed by one red. Hence

P (N = n) = (\frac{5}{7})^{n - 1} \frac{2}{7}, n = 1, 2, \dots .

This is a geometric distribution. Its total probability is one because the geometric series sums to

(2/7) / (1 - 5/7) = 1

.

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