Question

Three points $P,Q,R$ are selected at random from the circumference of a circle. The probability $p$ that the points lie on a semicircle is $\frac{k}{4}$.find $k$?

Answer 1

Let the arc length $\left(\overline{PQ}\right)=x;l\left(\overline{PR}\right)=y$ and let the length of circumference be 2s. (as shown in figure (i))
Clearly, $0\le x\le 2s;0\le y\le 2s\Rightarrow$ Total region $=2s\times 2s={4s}^2$
Now following four case arise which are favorable to
Case (i): $x,y\le s,$ then clearly the three points lie in semicircle
$\therefore x,y\in [0,s]$ is favorable region (as shown in figure (ii))
Case (ii): $x\le s,y\ge x+s$ (as shown in figure (iii))
Clearly $P,Q,R$ lie on semi-circle $x\in [0,s];y\in [x+s,2s]$ is favorable region
Case (iii) : $x\ge s,y\ge s$ clearly $P,Q,R$ lie on semi-circle with favorable region $x,y\in [s,2s]$ (as shown in figure (iv))
Case (iv): $y\le s;x\ge y+s$
$\therefore$ Favorable region is $y\in [0,s];x\in [y+s,2s]$ (as shown in figure (v))
The favorable region is
$\therefore$ $P\left(Event\right)$ $=\frac{s^2+\frac{1}{2}{s}^2+s^2+\frac{1}{2}{s}^2}{{4s}^2}=\frac{3}{4}$