 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-3/8h/c/solution

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

By stationarity,

P (X_{n - 2} = B ∣ X_{n} = B) = \frac{P ( X _{n - 2} = B , X _{n} = B )}{P ( X _{n} = B )} = \frac{π _{B} ( P ^{2} ) _{BB}}{π _{B}} = (P^{2})_{BB} .

The two-step return probability is

(P^{2})_{BB} = P_{B A} P_{A B} + P_{BB}^{2} + P_{BC} P_{CB} = \frac{3}{4} \frac{1}{4} + 0 + \frac{1}{4} \frac{1}{8} .

Therefore the requested probability is

\frac{7}{32} .

 New to topics? Read the docs here!