 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-1/19h/b/solution

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

By part (a) and repeated use of the Strong Markov property, the probability of descending from

(2, k)

(2, 0)

(\frac{2}{3})^{k} .

Starting from square 1 with £

m

, heads moves to

(2, m)

, whereas tails lands on square 3, loses £1, and moves to

(2, m - 1)

. Therefore the loss probability is

\frac{1}{2} (\frac{2}{3})^{m} + \frac{1}{2} (\frac{2}{3})^{m - 1} = \frac{5}{6} (\frac{2}{3})^{m - 1} .

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