Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 2 / 9F (Probability)

By the definition of conditional probability,Since the $B_j$ form a disjoint partition of the sample space, the law of total probability gives $\mathbb{P}(A)={\sum }_j\mathbb{P}(A\mid B_j)\mathbb{P}(B_j)$. Substitution proves Bayes' theorem in the required form:The hypotheses make the denominator positive and each conditioning probability well defined.

Let $N$ denote the mattress count and $U$ the event of undisturbed sleep. Use the intended model in which the presence of the pea is independent of $N$, with probability $1/2$, and without a pea sleep is undisturbed. Conditioning on $N$ givesThe law of total probability and the equal prior probabilities giveBy Bayes' theorem, the posterior probabilities of $N=6,7,8$ given $U$ are respectively $5/18,6/18,7/18$. Therefore the conditional expectation isUndisturbed sleep shifts the posterior toward larger mattress counts because those counts make the observation more likely.