Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 2 / 17 / c

The conditional probability product rule gives $P(A\mid B)P(B)=P(A\cap B)$. Applying it once more givesThe conditionals in the original expression require their conditioning events to have positive probability.

The events $B$ and $\overline{B}$ form a partition. The law of total probability therefore givesAs written, both conditionals require nonzero conditioning probabilities.

The numerator $P(B\mid A)P(A)$ equals $P(A\cap B)$, while $P(A\mid B)=P(A\cap B)/P(B)$. Thus, wherever the original quotient is defined,In particular a zero numerator and denominator cannot be cancelled; the original expression requires $P(A\mid B)>0$. This is the rearrangement of Bayes' theorem.

The two-event inclusion-exclusion principle subtracts the overlap counted twice:

This is exactly the definition of conditional probability, provided $P(A)>0$: