 Solution

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

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

The statement is true. For

n \geq 2

, being in every pairwise intersection is equivalent to being in all the events:

⋂_{i < j} (A_{i} \cap A_{j}) = ⋂_{k = 1}^{n} A_{k} .

Indeed every index occurs in at least one pair. Apply the union bound to the complements of these

(2 n)

pairwise intersections to obtain

P (⋂_{k = 1}^{n} A_{k}) \geq 1 - \sum_{i < j} {1 - P (A_{i} \cap A_{j})} = \sum_{i < j} P (A_{i} \cap A_{j}) - (2 n) + 1 > 0.

n = 2

, where it reduces to the asserted positivity of the single pairwise intersection.

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