 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/ia/paper-2/11f/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Conditioning on

U = X + Y

Z

gives

P (Z > u) = 1 - u

for

0 \leq u \leq 1

and zero for

u > 1

P (Z > X + Y) = \int_{0}^{1} (1 - u) u d u = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} .

 New to topics? Read the docs here!