BTGmoderatorLU wrote:Source: Manhattan Prep

Three points are chosen independently an at random on the circumference of a circle with radius r. What is the approximate probability that none of the three points lies more than a straight-line distance of r away from any other of the three points?

A. 1/9
B. 1/12
C. 1/18
D. 1/24
E. 1/27

The OA is B

Let the first point chosen be A, second point B and third point C. Point A can be chosen anywhere on the circle, because it doesn't matter where it is chosen.

Since B has to lie within 60 degrees of arc on either side of A, then B has a chance of (60 + 60)/360 = 1/3 of being chosen.
Now let's analyze where C could be. If A and B are 60 degrees apart, the minimum degree of freedom we can place on C is a 60-degree arc (i.e., C can be anywhere between A and B); in this case, the chance of placing C is 60/360 = 1/6. If A and B are the same point, then the maximum degree of freedom we can place on C is a 120-degree arc (i.e., C can be chosen just like B described in the previous paragraph); in this case, the chance of placing C is 120/360 = 1/3.
We don't know exactly where B is located in relation to A; however, as B moves farther away from A, up to a maximum of 60 degrees, the probability changes linearly (every degree it moves, adds one degree to where C could be). Therefore, we can average probabilities at each end to find (1/6 + 1/3)/2 = 1/4 as the chance we can place C based on a varying point B.
Therefore, the total probability of choosing points A, B and C where they don't lie more than a straight-line distance of r away from one another is 1 x 1/3 x 1/4 = 1/12.

Answer: B