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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-1/16/a/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 1 / 16 / a / Solution by

Codex  0 2026-10-18

Use the intended model of uniform draws and a replacement decision independent of the first ball's colour, with probability

r

. The marginal specification

P (R) = r

alone would not determine the experiment if replacement depended on that colour. This independence assumption is used throughout the question.

Write

\overline{B}_{j}

for a non-blue draw and

\overline{R}

for no replacement. The sample space consists of the eight ordered outcomes obtained by choosing one of

B_{1}, \overline{B}_{1}

, then one of

R, \overline{R}

, then one of

B_{2}, \overline{B}_{2}

. The chain rule for probabilities gives their probabilities:

outcome B_{1} \cap R \cap B_{2} B_{1} \cap R \cap \overline{B}_{2} \overline{B}_{1} \cap R \cap B_{2} \overline{B}_{1} \cap R \cap \overline{B}_{2} B_{1} \cap \overline{R} \cap B_{2} B_{1} \cap \overline{R} \cap \overline{B}_{2} \overline{B}_{1} \cap \overline{R} \cap B_{2} \overline{B}_{1} \cap \overline{R} \cap \overline{B}_{2} probability 4 r /25 6 r /25 6 r /25 9 r /25 (1 - r) /10 3 (1 - r) /10 3 (1 - r) /10 3 (1 - r) /10

For example, the first entry is

(2/5) r (2/5)

. The fifth is

(2/5) (1 - r) (1/4)

because one blue ball has been removed. With no replacement after a non-blue first draw, the two remaining colours each have two balls, explaining the last two entries. The first four entries sum to

r

and the last four to

1 - r

, so the full sample space has total probability one. At

r = 0

or

1

, the corresponding impossible outcomes simply have zero probability.

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