Question

If three vertices of a regular hexagon are chosen at random, then the chance that they form an equilateral triangle is

• A. $\frac{1}{3}$
• B. $\frac{1}{5}$
• C. $\frac{1}{10}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C

$\frac{1}{10}$

Explanation for the correct option:

Finding the probability of an equilateral triangle:

$\mathrm{ABCDEF}$ is a regular hexagon.

Total number of triangles ${=}^6{\mathrm{C}}_3$

$$\begin{gathered} =\frac{6!}{3!3!}\left[\because {}^n{C}_r=\frac{n!}{r!\left(n-r\right)!}\right] \\ =20 \end{gathered}$$

The only equilateral triangle possible are $∆\mathrm{ACE}$ and $∆\mathrm{BDF}$ from three vertices as shown in figure.

Therefore, the required probability is

$\begin{array}{rcl}& =& \frac{2}{{}^6{\mathrm{C}}_3}\\ & =& \frac{2}{20}\\ & =& \frac{1}{10}\end{array}$

Hence, the correct option is C.