Question

Three of six vertices of a regular hexagon are chosen at random.

The probability that the triangle with three vertices is equilateral equals

• A. $\frac{1}{2}$
• B. $\frac{1}{5}$
• C. $\frac{1}{10}$
• D. $\frac{1}{20}$

Answer 1

The correct option is C

$\frac{1}{10}$

Explanation for the correct answer:

Step 1: Find the number of total outcomes

There are $6$ vertices in a regular hexagon

The number of ways of choosing $3$ vertices of the $6$ is given by ${}^6{C}_3$

Step 2: List out the favorable outcomes

Consider the figure.

The only two equilateral triangles can be formed in a regular hexagon with the vertices,

$A,E,C$ and $D,E,F$

The triangles are equilateral as the lengths of the sides of these triangles, namely, $DF,FB,BD$ and $AE,EC,AC$ are equal and of length $\sqrt{3}x$,where,

$x$ is the length if the side of the regular hexagon.

Step 3: Use the classical definition of probability

Hence, the $\mathbf{Required}\mathbf{}\mathbf{probability}\mathbf{=}\frac{\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{favourable}\mathbf{}\mathbf{outcomes}}{\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{total}\mathbf{}\mathbf{outcomes}}$

$\mathrm{Required}\mathrm{probability}=\frac{2}{{}^6{C}_3}$

$\mathrm{Required}\mathrm{probability}=\frac{2}{{\displaystyle \frac{6!}{3!3!}}}$

$\mathrm{Required}\mathrm{probability}=\frac{2}{{\displaystyle 20}}=\frac{1}{10}$

Hence, the probability that a triangle chosen from the three vertices of a regular hexagon is equilateral is $\frac{1}{10}$.

Hence, option (C) is the correct answer.