 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-2/10f/c/solution

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

{1, 2, \dots}

P (X = m) = p (1 - p)^{m - 1} .

G_{X} (s) = \sum_{m = 1}^{\infty} p (1 - p)^{m - 1} s^{m} = \frac{p s}{1 - ( 1 - p ) s} .

 New to topics? Read the docs here!