2003/II/12
Planet Zog is a ball with centre $O$. Three spaceships $A$, $B$ and $C$ land at random on its surface, their positions being independent and each uniformly distributed on its surface. Calculate the probability density function of the angle $AOB$ formed by the lines $OA$ and $OB$.
Spaceships $A$ and $B$ can communicate directly by radio if $\angle AOB < \pi /2$, and similarly for spaceships $B$ and $C$ and spaceships $A$ and $C$. Given angle $\angle AOB =\gamma < \pi /2$, calculate the probability that $C$ can communicate directly with either $A$ or $B$. Given $\angle AOB =\gamma > \pi /2$, calculate the probability that $C$ can communicate directly with both $A$ and $B$. Hence, or otherwise, show that the probability that all three spaceships can keep in in touch (with, for example, $A$ communicating with $B$ via $C$ if necessary) is $( \pi +2)/(4\pi )$.
I leave you to figure out why the following are correct. For the first part
$P(\angle AOB \leq t) = \frac{(1-\cos t)2\pi}{4\pi}$, and hence p.d.f. is $f(t)=\tfrac{1}{2}\sin t$, where $0\leq t\leq \pi$.
Here is the answer to last part. It contains within it the answers to the middle parts.
$\int_{\gamma=0}^{\pi/2} (\frac{1}{2}+\frac{\gamma}{2\pi})f(\gamma)\,d\gamma+\int_{\gamma=\pi/2}^{\pi} (\frac{1}{4}-(\gamma-\frac{\pi}{2})\frac{1}{2\pi})f(\gamma)\, d\gamma= ( \pi +2)/(4\pi )$
Planet Zog is a ball with centre $O$. Three spaceships $A$, $B$ and $C$ land at random on its surface, their positions being independent and each uniformly distributed on its surface. Calculate the probability density function of the angle $AOB$ formed by the lines $OA$ and $OB$.
Spaceships $A$ and $B$ can communicate directly by radio if $\angle AOB < \pi /2$, and similarly for spaceships $B$ and $C$ and spaceships $A$ and $C$. Given angle $\angle AOB =\gamma < \pi /2$, calculate the probability that $C$ can communicate directly with either $A$ or $B$. Given $\angle AOB =\gamma > \pi /2$, calculate the probability that $C$ can communicate directly with both $A$ and $B$. Hence, or otherwise, show that the probability that all three spaceships can keep in in touch (with, for example, $A$ communicating with $B$ via $C$ if necessary) is $( \pi +2)/(4\pi )$.
I leave you to figure out why the following are correct. For the first part
$P(\angle AOB \leq t) = \frac{(1-\cos t)2\pi}{4\pi}$, and hence p.d.f. is $f(t)=\tfrac{1}{2}\sin t$, where $0\leq t\leq \pi$.
Here is the answer to last part. It contains within it the answers to the middle parts.
$\int_{\gamma=0}^{\pi/2} (\frac{1}{2}+\frac{\gamma}{2\pi})f(\gamma)\,d\gamma+\int_{\gamma=\pi/2}^{\pi} (\frac{1}{4}-(\gamma-\frac{\pi}{2})\frac{1}{2\pi})f(\gamma)\, d\gamma= ( \pi +2)/(4\pi )$
