Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 16

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: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.

For sampling without replacement, sum the two outcomes with a blue second draw:The marginal blue probability equals the initial blue fraction even though the two draws are dependent.

For sampling with replacement, the box has its original contents at the second draw. Therefore

Condition on whether replacement occurs. Both alternatives have blue second-draw probability $2/5$, so the law of total probability gives

The event of two blue draws splits into disjoint replacement and no-replacement cases. Their probabilities from the sample space are $4r/25$ and $(1-r)/10$. Hence

By the definition of conditional probability, or Bayes' theorem,Its denominator is positive throughout the specified range. Observing two blue balls raises the probability of replacement for $0<r<1$, because replacement makes a repeated colour more likely.

Let $C={\overline{B}}_1\cap B_2$. Under replacement its joint probability iswhereas without replacement it isIn the latter case the first removed ball is non-blue, so all $N_B$ blue balls remain. Applying Bayes' theorem and cancelling the positive common factors yields the posterior replacement probability after different-colour draws:It does not depend on $N_B$. To prove the requested bound directly,Equality holds at $r=0$ and $r=1$, and the inequality is strict between them. Its sketch joins $(0,0)$ to $(1,1)$, stays below the diagonal, and is increasing and convex becauseThe interpretation is that removing a non-blue ball increases the chance of blue on the second draw; seeing the two different colours therefore favours the no-replacement explanation.