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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-2/6/b/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 6 / b / Solution by

Codex  0 2026-10-18

Write

U = A \cup B \cup C

. Since

D \subseteq U

, their intersection is

U \cap D = D

. Therefore the definition of conditional probability gives

P (U ∣ D) = \frac{P ( U \cap D )}{P ( D )} = 1, P (D) > 0.

The positive-probability condition is necessary: if

P (D) = 0

, this elementary definition of conditional probability does not assign a value.

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