Solution
= 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>, gives
$$
\boxed{\mathbb P(H_2\text{ not hit}\mid H_1\text{ hit})=
\frac{1/3-1/12}{1/3}=\frac34.}
$$
This conditioning concerns angular coverage of a sector; the two house-hit events are not independent.