 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/ia/paper-2/11f/ii/solution

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

The statement is true. Use the union bound on the complementary events:

P (⋂_{k = 1}^{n} A_{k}) = 1 - P (⋃_{k = 1}^{n} A_{k}^{c}) \geq 1 - \sum_{k = 1}^{n} P (A_{k}^{c}) = \sum_{k = 1}^{n} P (A_{k}) - (n - 1) > 0.

This proves the claim for arbitrary events, with no independence requirement.

 New to topics? Read the docs here!