1. Total number of ways to choose 3 vertices from 6

From the 6 vertices of a regular hexagon, the number of ways to choose any 3 vertices is

$\binom{6}{3} = 20.$

2. Condition for an equilateral triangle

We need to count how many triangles formed by choosing 3 vertices of the regular hexagon are equilateral.

In a regular hexagon, all vertices lie on a circle, equally spaced at angles of $60^\circ$.

For a triangle formed by vertices of this hexagon to be equilateral, the chosen vertices must be equally spaced around the circle.

So we must select every second vertex.

The possible selections are:

• $(1,3,5)$
• $(2,4,6)$

These are the only two such triangles.

Hence, the number of favorable cases is

$2.$

3. Compute the probability

$P(\text{equilateral triangle}) = \frac{\text{favorable cases}}{\text{total cases}} = \frac{2}{20} = \frac{1}{10}.$

4. Match with the options

$\frac{1}{10}$ corresponds to Option A.

Final Answer

$\boxed{\frac{1}{10}}$