 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/ib/paper-2/20h/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

The states

b, d

p_{n} = P (X_{n} = b ∣ X_{0} = d)

. Within this class,

p_{n + 1} = \frac{1}{4} p_{n} + \frac{2}{3} (1 - p_{n}) = \frac{2}{3} - \frac{5}{12} p_{n}, p_{0} = 0.

8/17

. Subtracting that fixed point and iterating gives the exact transition probability

p_{n} = \frac{8}{17} (1 - (- \frac{5}{12})^{n}), n \geq 0.

 New to topics? Read the docs here!