 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/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 .

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

 New to topics? Read the docs here!