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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2021/discrete-mathematics-continuous-mathematics-and-probability/8/a/v/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Discrete mathematics, continuous mathematics, and probability / 8 / a / v / Solution by

Codex  0 2026-10-25

Conditional on a contiguous red block, its

n + 1

possible starting positions remain equally likely because their original colour patterns had equal probabilities. The first card is red exactly when that block starts at position

1

. Hence

P (first red ∣ all reds together) = \frac{1}{n + 1} .

(1)

This also follows by dividing the probability of the single favourable block position,

1/ (m m + n)

, by the probability of all

n + 1

block positions.

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