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Three of the six vertices of a regular h...

Three of the six vertices of a regular hexagon are chosen the random. What is the probability that the triangle with these vertices is equilateral.

A

`(1)/(10)`

B

`(1)/(5)`

C

`(3)/(10)`

D

`(3)/(20)`

Text Solution

Verified by Experts

The correct Answer is:
A
Since, there is a regular hexagon, then the number of ways of choosing three vertices is `^6C_3` And, there is only two ways i.e. choosing vertices of a regular hexagon alternate, here`A_1,A_3,A_5 or A_2,A_4,A_6` wiil result in an equilateral triangle.
`therefore" Required probility

`therefore` Required pobability
`=2/(""^(6)C_3)=2/((6!)/(3!3!))=(2xx3xx2xx3xx2)/(6xx5xx4xx3xx2xx1)=1/10`
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