Question

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

• A. 44563
• B. 44566
• C. 44571
• D. 44581

Answer 1

Answer: (C)

44571

Answer 2

Three vertices out of 6 can be chosen in $^6C_3$ways So, total ways =$^6C_3 =20$ Only two equilateral triangles can be formed $\Delta$AEC and $\Delta$BFD $\therefore \, \, \, \, \,$Favourable ways = 2 So, required probability = $\frac{2}{20}=\frac{1}{10}$