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  4. Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with three vertices is equilateral, equals

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Q. Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with three vertices is equilateral, equals

Probability

Solution:

Three vertices out of $6$ can be chosen in $^{6}C_{3}$ ways
So, total ways $=^{6}C_{3}=20$
image
only two equilateral triangles can be formed $\Delta\,AEC$ and $\Delta\,BFD$
So, favourable ways = $2$.
$\therefore $ required probability $=\frac{2}{20}=\frac{1}{10}=0.1$