 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-2/20h/c/solution

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

Let

v_{i}

be the vertex

i

steps from

B

toward

C

, and let

x_{i}

be the mean time to reach

v_{i + 1}

from

v_{i}

. Part (b) gives

x_{0} = 4 n + 1

. For

1 \leq i < n

v_{i}

gives

x_{i} = \frac{1}{2} \cdot 1 + \frac{1}{2} (1 + x_{i - 1} + x_{i}),

because a step to

v_{i - 1}

must be followed by a fresh passage of mean

x_{i - 1}

back to

v_{i}

before trying again. Hence

x_{i} = x_{i - 1} + 2 = 4 n + 1 + 2 i .

By the Strong Markov property, the expected time from

B

C

is the sum of these successive passage times:

E_{B} [T_{C}] = \sum_{i = 0}^{n - 1} x_{i} = n (4 n + 1) + n (n - 1) = 5 n^{2} .

Adding the initial mean time

n^{2}

from

A

B

gives

E_{A} [T_{C}] = 6 n^{2} .

 New to topics? Read the docs here!