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:

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TestHub · Accepted Answer

Total number of ways of selecting three points from vertices of a regular hexagon

$= {}^{6}{C_{3}} = \frac{6.5 . 4}{1.2 . 3} = 20$
Favorable selection (forming equilateral triangle)
$= 2 (ACE, BDF)$
$\Rightarrow$ Required probability $P = \frac{N u m b e r o f f a v o u r a b l e s e l e c t i o n s}{T o t a l s e l e c t i o n s}$
$= \frac{2}{20} = \frac{1}{10}$

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