Honestly it took me a while to even understand the solution. If anyone got this problem on the GMAT, I would consider him or her extremely unlucky.
Consider the 3 points to be A,B and C.
The first point 'A' could be chosen anywhere in the circle, no restrictions.
I am borrowing an image posted by Mitch for a different problem.
https://postimage.org/image/11xb37dyc/
Now point B could be chosen anywhere such that the
distance from A is between 0 to r
The corresponding probability for the distance (0-r) would be from
(0 to 1/3) (0-120 degrees)
Now if point A and B were close to each other (together), point C could similarly be chosen between distance r on either sides or
P(1/3)
If B were at a distance 'r' from A, C should be between A and B on the circumference (or between the sector of 60 degree) Prob will be 1/6.
Together we know that the probability will be in-between
1*1/3*1/3 (to) 1*1/3*1/6 (For points A,B and C correspondingly)
(or)
1/9 to 1/18.
Only
B satisfies this. I doubt theres even a finite probability for this. I don't think I can explain the solution well enough for anyone to understand, unfortunately.
