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3 out of 6 vertices of a regular hexagon are chosen at a time at random. The probability that the triangle formedwith these three vertices is an equilateral triangle, is
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$\frac{1}{10}$
Give that, 3 out of 6 vertices of hexagon are chosen.
Total number of outcomes $={ }^5 \mathrm{C}_3=\frac{6 !}{3 ! 3 !}=20$
Only two equilateral triangles are possible of regular hexagon, i.e. $\triangle \mathrm{AEC}$ and $\triangle \mathrm{BFD}$.
$\therefore \quad \text { Probability }=\frac{2}{20}=\frac{1}{10}$
Total number of outcomes $={ }^5 \mathrm{C}_3=\frac{6 !}{3 ! 3 !}=20$
Only two equilateral triangles are possible of regular hexagon, i.e. $\triangle \mathrm{AEC}$ and $\triangle \mathrm{BFD}$.
$\therefore \quad \text { Probability }=\frac{2}{20}=\frac{1}{10}$
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