The probability that all the N points lie on a semi-circle is the same as the probability that the rest $(n-1)$ points lie on an arc of width $\pi$ in the clockwise direction from the first point which can be $1,2, ..., N$. Now, since the individual events are exclusive, we have,
A nice solution was proposed by Raghav Agrawal -
ReplyDeleteThe probability that all the N points lie on a semi-circle is the same as the probability that the rest $(n-1)$ points lie on an arc of width $\pi$ in the clockwise direction from the first point which can be $1,2, ..., N$. Now, since the individual events are exclusive, we have,
$
P = n \times \frac{1}{2^{n-1}}
$