 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/5/e/i/solution

Each visit to

I

corresponds to two steps, one out of

B

and one out of

I

. Hence a successful path with

n

visits contains

2 n

steps. Taking its expectation over the geometric distribution gives

S_{A B} = n = 1 \sum \infty 2 n q^{n - 1} p .

The last pair is

B \to I \to A

; it contributes two steps just like each previous

B \to I \to B

excursion.

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