Tough!!!!!!

the first point can be located anywhere. Now lets call it A.
Now from A, the maximum distance can be 2r and minimum can be 0. Also R distance means 120 degree at centre or 1/3 prob of choosing B.
now if B is the extreme distant point, then 1/6 for C and if B is the nearest point, 1/3 for C
average: 1/3 + 1/6 / 2 = 1/4
thus 1/3*1/4=1/12

Hi cans,

Could you explain why you took the average of the probabilities for choosing point C? Please explain the logic behind taking the average.

//sungoal

Hi cans,

Could you explain why you took the average of the probabilities for choosing point C? Please explain the logic behind taking the average.

//sungoal

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.

This is NOT a question that would ever appear on the GMAT.