 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-2/11f/i/solution

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

Let

B_{n}

count the

+ 1

steps. Then

B_{n} \sim Binomial (n, p), S_{n} = B_{n} - (n - B_{n}) = 2 B_{n} - n .

Consequently

P (S_{n} = 2 k - n) = (k n) p^{k} q^{n - k}, k = 0, \dots, n,

and

S_{n}

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