Probability-5

Total number of ways of picking 3 vertices = 6C3= 20 Only two sets of 3 vertices will have an equilateral triangle.

So probability = 2/20 = 1/10

First, all possible choices

6 choices of the first vertex, leaving 5 choices of the second, and 4 choices of the third: 6(5)(4).

Now, ways of choosing vertices that form a hexagon

As before, 6 choices of the first vertex. But 2 choices of the next vertex. Having chosen the second vertex there is only a single possibility for the third. Then all ways of choosing: 6(2).

Overall probability

(6(2))/(6(5)(4)) = 1/10

Let the vertices be numbered as 1,2,3,4,5,6

Then total ways of selection =20

Total number of equilateral triangles=2 (1,3,5 and 2,4,6)

Hence Ans- 1/10