 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ia/paper-2/9f/a/solution

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

p

{1, 2, \dots}

P (X = n) = (1 - p)^{n - 1} p, n \geq 1.

It counts the trial on which the first success occurs in independent trials with success probability

p

p s / [1 - (1 - p) s]

. Differentiating at

s = 1

gives

EX = \frac{1}{p}, Var X = \frac{1 - p}{p ^{2}} .

This support convention is important: a count of failures before the first success would start at zero.

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