 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-2/11f/b/solution

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

Conditioned on exactly

k

heads, every

k

-element set of head positions is equally likely. There are

(k n)

such sets. A run containing all

k

heads can begin at any of the positions

1, 2, \dots, n - k + 1,

and each beginning determines one admissible set. Thus, for

k \geq 1

P (the k heads are consecutive ∣ exactly k heads) = \frac{n - k + 1}{( k n )} .

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