Triangle in Hexagon

three vertices can be chosen in 6C3 ways (order is not important). Means we can have 6C3 = 20 different triangles by choosing 3 vertices of regular hexagon. and we can have equilateral triangles only in two ways as shown in image below.

hexagons

so the required probability is 2/20 = 1/10

Given that you pick a certain vertex, you know exactly the other two points you need to pick to form an equilateral triangle. For the second vertex, you are picking one of $2$ points out of $5$ possible points. Then you have to pick $1$ vertex out of $4$ remaining to create the triangle.

The answer is $\dfrac{2}{5}\times\dfrac{1}{4}=\boxed{\dfrac{1}{10}}$

3 vertices from 6 can be selected as 6C2 and only 2 triangle will form equilateral on joining alternate corners of hexagon so its 2/20 which is 1/10...

To construct an equilateral triangle, we need to select alternating vertices. The first vertex can be chosen in 6 ways. The second vertex can only be chosen in 2 ways. And third can only be picked in 1 way. So we have $6 \times 2 \times 1$ ways of picking an equilateral triangle.

Number of ways to pick any three vertex, using pigeonhole principle is $6 \times 5 \times 4$

Therefore the probability is $\frac { 6 \times 2 \times 1}{6 \times 5 \times 4} = \frac {1}{10}$

Ration of (there are two equilateral triangle with joining alternate three vertices) to total triangle formed by its 6 vertices is 6C3= 2/20=1/10

Let's choose them in order. It doesn't matter which we choose first.

Out of the 5 remaining, only two of them can be part of an equilateral triangle. So 2/5.

Out of the 4 remaining, only one of them can form an equilateral triangle if we chose one of the right ones before. So 1/4.

Thus the answer is 2/5 x 1/4 = 2/20 = 1/10

the vertices can be chosen 6c3 ways.equilateral triangle can b 2.