 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-4/7h/a/solution

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

Let

g (0) = 0

g (1) = 1 + \frac{1}{2} g (2),

g (2) = 1 + \frac{1}{2} g (1) + \frac{1}{2} g (3),

and, because state

3

moves to

2

or remains at

3

g (3) = 1 + \frac{1}{2} g (2) + \frac{1}{2} g (3) .

The last equation gives

g (3) = 2 + g (2)

. Substitution into the second gives

g (2) = 4 + g (1)

, and the first then gives

g (1) = 3 + g (1) /2

. Hence

g (1) = 6, g (2) = 10, g (3) = 12 .

These are the expected hitting times of the absorbing state.

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