Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 2 / 9F (Probability)

The Kolmogorov axioms for a probability measure $\mathbb{P}$ are nonnegativity, normalization $\mathbb{P}(\Omega )=1$, and countable additivity on pairwise disjoint events. For $\mathbb{P}(B)>0$, conditional probability is defined byBecause the events $A\cap B_i$ are pairwise disjoint and their union is $A$, countable additivity gives the law of total probabilityFor any $j$ with $\mathbb{P}(A)>0$, the definition of conditional probability and this identity give Bayes theorem:

The prior probabilities of choosing the two urns are equal. Their white-ball probabilities areBy Bayes theorem,

The two possible ways to observe one ball of each colour have probabilitiesConditioning on their union therefore gives

Urn $i$ has $2^i$ balls, so a draw from it is white with probability $2^{-i}$. Given that the first ball is white, Bayes theorem gives the posterior probabilityAfter replacement, the second draw is conditionally independent of the first given the urn. HenceConsequently $\boxed{p(n)\to 1/3}$.