Three of six vertices of a regular hexagon are chosen at random. The probability that the triangle with three vertices is equilateral equal to :
Step-by-Step Solution
Key Concept: In a regular hexagon with vertices labeled 0-5 in order, an equilateral triangle is formed only when vertices are separated by equal angular intervals of 120°, corresponding to alternating vertices (like vertices {0,2,4} or {1,3,5}). The total number of ways to choose 3 vertices from 6 is C(6,3)=20, and exactly 2 equilateral triangles exist, giving probability 2/20 = 1/10.
Total ways to choose 5 points from 5 collinear points is $^5C_5 = 20$. For an equilateral triangle, all three vertices must be non-collinear, but choosing from 5 collinear points permits only degenerate triangles. Only 2 configurations yield an equilateral triangle. Required probability is $\frac{2}{20} = \frac{1}{10}$.
Correct Answer: 3