M7MBA wrote:n points are equally spaced on a circle, where n is an even number greater than 3. If 3 of the n points are to be chosen at random, what is the probability that a triangle having the 3 points chosen as vertices will be a right triangle?

\(A. \dfrac{n−1}{6}\)
\(B. \dfrac{n+2}{6}\)
\(C. \dfrac{2}{3n+2}\)
\(D. \dfrac{3}{n−1}\)
\(E. \dfrac{6}{n+4}\)

The OA is the option _D_.

Source: Manhattan GMAT

We can let n = 4; we see that any 3 points we pick will form a right triangle. So the probability is 1. We see that choices B and D both yield 1 when n = 4. However, if n = 6, then choice B will be a number greater than 1, which is not possible for a number representing a probability. Therefore, the correct answer must be D.

Answer: D

I did this the same way Scott did - with those answer choices, it's a very fast question. I doubt they meant to make the question so easy to answer.

If we had better-chosen answer choices, and needed to actually solve: if you connect three points on a circle, you only get a right triangle if one edge of the triangle is a diameter. When we have an even number of equally spaced points around a circle, each point will be diametrically opposite another point. We can reverse the question, and work out first the probability we don't pick two diametrically opposite points when we pick three of the points around the circle. It doesn't matter which point we pick first, but when we pick the second point, we have n-1 points left, and we must avoid the point opposite our first point, so the probability we don't pick a diameter is (n-2)/(n-1). Then when we pick the third point, we have n-2 points left, and we must avoid the two points diametrically opposite our first two selections, so the probability we don't create a diameter is (n-4)/(n-2). Multiplying these, the probability we don't make a right triangle is

[ (n-2) / (n-1) ] * [ (n-4) / (n-2) ] = (n-4) / (n-1)

and subtracting that from 1, the probability we do make a right triangle is

1 - [ (n-4) / (n-1) ] = [(n-1) - (n-4)] / (n-1) = 3/(n-1)