 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ia/paper-2/4f/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

S_{n} = \sum I_{i}

, each

E I_{i} = 1/ n

. Distinct indicators have joint probability

1/ [n (n - 1)]

and zero covariance, so

E S_{n} = 1

and

var S_{n} = n (1/ n) (1 - 1/ n) = 1 - 1/ n

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