Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 2 / 12F / b / iii / Solution

The event that $H_1$ is hit but $H_2$ is not has probability $1/3-1/12=1/4$. Dividing by the probability $1/3$ that $H_1$ is covered, using the definition of conditional probability, givesThis conditioning concerns angular coverage of a sector; the two house-hit events are not independent.