Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these three vertices is equilateral equals :

Options

A. a. `1/6`
B. b. `1/3`
C. c. `1/10`
D. d. none of these

Answer

To solve the problem, we need to find the probability that a triangle formed by randomly selecting three vertices of a regular hexagon is equilateral.

### Step-by-Step Solution:

1. **Identify the Total Number of Vertices:**
A regular hexagon has 6 vertices.

2. **Calculate the Total Number of Ways to Choose 3 Vertices:**
The total number of ways to choose 3 vertices from 6 is given by the combination formula:
\[ \text{Total combinations} = \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

3. **Identify the Equilateral Triangles:**
In a regular hexagon, the equilateral triangles that can be formed by choosing 3 vertices are:
- Triangle formed by vertices 1, 3, and 5 (A, C, E)
- Triangle formed by vertices 2, 4, and 6 (B, D, F)

Thus, there are 2 equilateral triangles.

4. **Calculate the Probability:**
The probability \( P \) that a randomly chosen triangle is equilateral is given by the ratio of the number of favorable outcomes (equilateral triangles) to the total outcomes (all triangles):
\[ P = \frac{\text{Number of equilateral triangles}}{\text{Total combinations}} = \frac{2}{20} = \frac{1}{10} \]

### Final Answer:
The probability that the triangle formed by choosing 3 vertices of a regular hexagon is equilateral is \( \frac{1}{10} \).

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