Three of six vertices of a regular hexagon are chosen at random. The probability that the triangle with…
Three of six vertices of a regular hexagon are chosen at random. The probability that the triangle with these three vertices is equilateral, equals
$\frac{1}{2}$
$\frac{1}{5}$
$\frac{1}{10}$
$\frac{1}{20}$
Solution
The number of triangles that can be drawn using 6 vertices is given by
$\mathrm{n}(\mathrm{S})={ }^6 \mathrm{C}_3=20$
$A$ : Event of selecting equilateral triangle. The equilateral triangle can be drawn if selected three vertices are alternate.
$\begin{aligned} & \therefore \quad \mathrm{n}(\mathrm{A})=2 \\ & \therefore \quad \mathrm{P}(\mathrm{A})=\frac{2}{20}=\frac{1}{10}\end{aligned}$