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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/2016/paper-2/16/a/iv/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 / 2016 / Paper 2 / 16 / a / iv / Solution by

Codex  0 2026-10-18

Three draws cannot all be green because there are only two green balls. Thus the probability of at least one red ball is $1$ . At least two red balls means

X \leq 1

, so

P (at least two red) = \frac{1}{10} + \frac{3}{5} = \frac{7}{10} .

(1)

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