 Solution

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

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

G_{X} (s) = \sum_{m = 1}^{\infty} P (X = m) s^{m} .

Differentiating

n

G_{X}^{(n)} (s) = \sum_{m = n}^{\infty} \frac{m !}{( m - n )!} P (X = m) s^{m - n} .

s = 0

, only the term

m = n

remains. Hence

P (X = n) = \frac{G _{X}^{(n)} ( 0 )}{n !} .

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