Combinations, Probability

[swerve]

A semicircle with area of \(x \pi\) is marked by seven points equally spaced along the half arc of the semicircle, such that two of the seven points form the endpoints of the diameter. What is the probability of forming a triangle with an area less than \(x\) from the total number of triangles formed by combining two of the seven points and the center of the diameter?

A. \(4/5\)
B. \(6/7\)
C. \(17/21\)
D. \(19/21\)
E. \(31/35\)

The OA is A

[regor60]

A little work can eliminate what otherwise could be a lot of work.

The question stipulates that the center is one point of the triangle so that means 2 points must be selected from the 7 to potentially form a triangle. The number of ways to select 2 points from the 7 is:

7!/2!5! = 21

Recognize that in order to form a triangle the 3 points must not fall on a line.

One set of 3 points, the center and the two points forming the diameter, are on a line. So this single set must be subtracted from the 21:

21 - 1= 20 triangles.

Examining the answer choices , only [spoiler]A, 4/5, [/spoiler] has a denominator that can be arrived at from 20