Probability of 3 points on a circle containg the center
Answers
Mathematics: If three points are chosen at random on a circle, and a triangle is formed by taking the three chosen points as its vertices, what is the probability that the triangle thus formed contains the center of the circle within it?
Let’s take three points A, B, C on the circumference of a circle (with radius R), which construct a triangle. Take two points (say A and B) on the circle which are separated by x (in radian) angular distance from each other. Therefore the third point (C) has to be within an arc length of R.x such that the center of the circle lies within the triangle (see the diagram attached).
The probability of the point C to lie within the arc length is= (R*x)/ (2* pi*R)=x/(2*pi).
But the points A and B can lie any where within the arc of the circle i.e. the angle x may vary arbitrarily. Therefore we have to average out the above probability within the allowable domain of x.
Therefore the final probability is= 2* Integrate[(x/(2*pi)), {dx, 0, pi}]
= 1/4 (Ans)