Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 2 / 12F (Probability)

Normalization of the probability density function givesThe first two moments areConsequentlyThe cumulative distribution function is $F(\varphi )=\varphi /\pi -{\varphi }^2/(4{\pi }^2)=1-(1-\varphi /(2\pi ){)}^2$ on $[0,2\pi ]$. For $U$ with the uniform distribution on $(0,1)$, inverse transform sampling therefore usesThe inverse lies in the required interval and satisfies $F(\Phi )=U$.

Conditional on $\Phi =\varphi$, the uniform distribution of the sector's centre angle gives a range of possible centre bearings of length $\varphi$ that hit a specified house. Independence of the two angles therefore gives the conditional probability $\varphi /(2\pi )$. By the law of total expectation,

For coverage of antipodal points by a randomly oriented sector, the house bearings differ by $\pi$. For $\varphi <\pi$, no single sector can cover both. For $\varphi \ge \pi$, the two allowable arcs of centre bearings overlap in two components, each of length $\varphi -\pi$. Thus the conditional probability of a double hit is $(\varphi -\pi )/\pi$, not $(\varphi -\pi )/(2\pi )$. Averaging against the probability density function givesPut $v=\varphi -\pi$ to obtain $\frac{1}{2{\pi }^3}{\int }_0^{\pi }v(\pi -v)dv$. Therefore

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.