Probability

[swerve]

\(n\) points are equal 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. \(6/(n+4)\)

The OA is D

Source: Manhattan Prep

[Scott@TargetTestPrep]

\(n\) points are equal 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. \(6/(n+4)\)

The OA is D

Solution:

When there are an even number of equally spaced points on a circle, the points are actually the endpoints of the diameters of the circle. Recall that the angle subtended by the diameter of a circle is always a right angle.

Let the first point be chosen. If the second point is the other endpoint of the diameter containing the first point, then no matter which point is the third point, the triangle will be a right triangle. The probability of this is 1/(n - 1) since there are n - 1 points besides the first point and only one of them is favorable.

If the second point is not the other endpoint of the diameter containing the first point (an event with a probability of (n - 2)/(n - 1)), then the third point can either form a diameter with the first point or the second point. Thus, of the remaining n - 2 points, two of them are favorable and the probability is 2/(n - 2). The probability of this scenario is (n - 2)/(n - 1) x 2/(n - 2) = 2/(n - 1).

Thus, the probability that the three points form a right triangle is 1/(n - 1) + 2/(n - 1) = 3/(n - 1).

Answer: D