Given a semi circle, the probability that a uniform random point will lie on that semi circle is $\frac{1}{2}$. If three points lie on a semi circle then they definitely lie on a semicircle starting at exactly one of the points. There are 3 points, thus three semi circles and the events that the all the points lie on a semi circle starting at each of the points are mutually exclusive. Therefore the probability is $3\times\frac{1}{2^2}$. This is easily extended to $n$ points as $\frac{n}{2^{n-1}}$.
P.S: I read the solution on [this](https://sbjoshi.wordpress.com/2009/10/04/puzzle-semi-circle-covering-n-points/) blog by Saurabh Joshi
Given a semi circle, the probability that a uniform random point will lie on that semi circle is $\frac{1}{2}$. If three points lie on a semi circle then they definitely lie on a semicircle starting at exactly one of the points. There are 3 points, thus three semi circles and the events that the all the points lie on a semi circle starting at each of the points are mutually exclusive. Therefore the probability is $3\times\frac{1}{2^2}$. This is easily extended to $n$ points as $\frac{n}{2^{n-1}}$.
ReplyDeleteP.S: I read the solution on [this](https://sbjoshi.wordpress.com/2009/10/04/puzzle-semi-circle-covering-n-points/) blog by Saurabh Joshi