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Question
Three of the six vertices of a regular hexagon are chosen at random. What is the probability that the triangle with these vertices is equilateral.
Solution
We choose three vertices out of 6 in 6C3 = 20 ways.
∴ Total number of elementary events = n(S) = 20
Number of ways of choosing an equilateral triangle = 2
i.e. A1A3A5 or A2A4A6
∴ Total number of favourable events = n(E) = 2
Hence, required probability = \[\frac{2}{20} = \frac{1}{10}\]
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