Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 2 / 9E (Probability)

For $\mathbb{P}(B)>0$, the conditional probability isHenceBecause the $B_k$ form a partition, the law of total probability givesSubstitution proves Bayes theorem in the required form:

Urn $r$ has $n-r$ blue balls among $n-1$ balls. Averaging over the uniformly chosen urn,The probability that both removed balls are blue isThereforeThe first blue draw makes urns with more blue balls more likely, which explains why the conditional probability exceeds one half.

Events $A$ and $B$ are independent whenFor a possible sum $2\le t\le 12$, let $N_t$ be the number of ordered die pairs with sum $t$. ThenIf $t-i\in \{1,\dots ,6\}$, then $\mathbb{P}(A_t\cap B_i)=1/36$; otherwise it is zero. Independence for a possible sum therefore requiresso $N_t=6$. This occurs only for $t=7$, and then $7-i$ is valid for every $i=1,\dots ,6$. ThusIf impossible sums are admitted as events of probability zero, they are trivially independent as well.

Call the coins $1,2,3$ in the order given. Coin 1 beats the constant score of coin 2 exactly when it shows a head, so coin 1 wins with probability $3/5$. Coin 2 beats coin 3 exactly when coin 3 shows a head and scores three, again with probability $3/5$.

Coin 3 beats coin 1 whenever coin 3 shows a tail, or when coin 3 shows a head and coin 1 shows a tail. Its winning probability is thereforeThus the preferences form a nontransitive cycle:The second chooser can always select a coin that has winning probability greater than one half against the first choice. Therefore