If three of six vertices of a regular hexagon are chosen at random, then probability that triangle f

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14.Probability

hard

If three of the six vertices of a regular hexagon are chosen at random, then the probability that the triangle formed with these chosen vertices is equilateral is

A

$\frac{3}{{10}}$

B

$\frac{1}{{5}}$

C

$\frac{1}{{10}}$

D

$\frac{3}{{20}}$

(JEE MAIN-2019)

Solution

Only two equilateral triangle are possible $A_{1} A_{3} A_{5}$ and $A_{2} A_{5} A_{6}$

$\frac{2}{^{6} \mathrm{C}_{3}}=\frac{2}{20}=\frac{1}{10}$

Standard 11

Mathematics

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