n points are equally spaced on a circle, where n is an even

[BTGmoderatorDC]

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. (n−1)/6(n−1)/6

B. (n+2)/6(n+2)/6

C. 2/(3n+2)2/(3n+2)

D. 3/(n−1)3/(n−1)

E. 6/(n+4)

OA D

Source: Manhattan Prep

[fskilnik@GMATH]

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. (n−1)/6

B. (n+2)/6

C. 2/(3n+2)

D. 3/(n−1)

E. 6/(n+4)

Source: Manhattan Prep

Let´s explore the simplest particular case: n = 4

Choosing any 3 points among A, B, C, D (see figure), it´s CERTAIN that we will have chosen a right triangle, therefore when n = 4 we expect 1 as our "target".


(A) 3/6 is not 1, refuted
(B) 6/6 is 1, this alternative is a "survivor"
(C) 2/14 is not 1, refuted
(D) 3/3 is 1, this alternative is a "survivor"
(E) 6/8 is not 1, refuted

Now let´s compare (B) (n+2)/6 and (D) 3/(n-1) ...

When n increases, (n+2)/6 increases and this is not good, hence (B) is refuted. (Reason: with greater values of n, the probability of getting a right triangle decreases!)

The only "survivor" (D) is the right alternative choice!


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.


POST-MORTEM: (Solving the general case to prove that (D) is really the right answer, without excluding the other choices, as we did before.)
There are C(n,3)*3! = n(n-1)(n-2) equiprobable possible choices if we consider the order in which the points are chosen in the circle (to help the "favorable counting", below)!
A favorable situation is such that 2 of the 3 chosen points must be opposite to each other (to be a diameter of the circle).
First choice : n possibilities ("free") , second choice (scenario A): 1 possibility (to have the diameter already) and third choice: (n-2) choices (any point remaining is good) :: n.1.(n-2) choices
First choice : n possibilities ("free"), second choice (scenario B): (n-2) possibilities (NOT the diameter yet) and third choice: 2 choices (diameter with first or second choice) :: n.(n-2).2 choices
Scenarios A and B are mutually exclusive, hence we may add their numbers of possibilities: 3n(n-2) choices
? = 3n(n-2) divided by C(n,3)*3! equals 3/(n-1) , the expected answer!