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}}