 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-2/10e/a/solution

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

Ignore rolls showing

2, 3, 4, 5

and consider the first roll among

{1, 6}

A

that a 1 occurs before a 6 has probability

1/2

. The event that the very first roll is 1 has probability

1/6

and is contained in

A

. Hence

P (first roll is 1 ∣ A) = \frac{1/6}{1/2} = \frac{1}{3} .

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