Three of six vertices of a regular hexagon are chosen at random. Tie probability that the triangle formed by these vertices is an equilateral triangle is

Options

A. `3/20`
B. `3/10`
C. `(-1)/10`
D. `1/4`

Answer

To find the probability that a triangle formed by randomly choosing three vertices of a regular hexagon is an equilateral triangle, we can follow these steps:

### Step 1: Identify the total number of ways to choose 3 vertices from 6 vertices of a hexagon.
The total number of ways to choose 3 vertices from 6 can be calculated using the combination formula:

\[ \text{Total ways} = \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

### Step 2: Determine the number of favorable outcomes (equilateral triangles).
In a regular hexagon, the vertices can be labeled as A, B, C, D, E, and F. The equilateral triangles that can be formed by choosing 3 vertices are:
1. Triangle A, C, E
2. Triangle B, D, F

Thus, there are a total of 2 equilateral triangles.

### Step 3: Calculate the probability.
The probability \( P \) of forming an equilateral triangle is given by the ratio of the number of favorable outcomes to the total outcomes:

\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{2}{20} = \frac{1}{10} \]

### Final Answer:
The probability that the triangle formed by the chosen vertices is an equilateral triangle is \( \frac{1}{10} \).

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