 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-1/20h/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Let

T_{k}

be the expected capture time starting from

D_{0} = k

, so

T_{0} = 0

. The Strong Markov property at successive first passages to lower levels and part (a) give

T_{k} = T_{1} + (k - 1) μ, k \geq 1.

D_{n} = 1

, the next state is zero with probability

r /2

1/2

, and two with probability

(1 - r) /2

T_{1} = 1 + \frac{1}{2} T_{1} + \frac{1 - r}{2} T_{2},

T_{1} = 2 + (1 - r) (T_{1} + μ) .

Solving,

T_{1} = \frac{2}{r} + (\frac{1}{r} - 1) μ .

Substituting

μ = 2/ (2 q - 1)

now yields

T_{m} = \frac{2}{r} + (m + \frac{1}{r} - 2) \frac{2}{2 q - 1},

which is the required expected capture time.

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