Past exam of the mathematics course of the University of Cambridge / 2013 / ia / Paper 2 / 12F / Solution

By the partition and the law of total probability, $\mathbb{P}(A)={\sum }_j\mathbb{P}(A\cap B_j)={\sum }_j\mathbb{P}(A\mid B_j)\mathbb{P}(B_j)$. Also $\mathbb{P}(B_i\mid A)=\mathbb{P}(A\cap B_i)/\mathbb{P}(A)$. Combining these identities proves Bayes theorem:The assumed positive probabilities make the conditional expressions well defined.