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 :

Question

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 : (A) (3/10) (B) (1/10) (C) (3/20) (D) (1/5)

Tardigrade Admin · Accepted Answer

Only two equilateral tringles are possible A1 A3 A5 and A2A5A6 (2/6C3) = (2/20) = (1/10)

Q. 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 :

$\frac{3}{10}$

$\frac{1}{10}$

$\frac{3}{20}$

$\frac{1}{5}$

Only two equilateral tringles are possible $A_{1} A_{3} A_{5}$ and $A_{2} A_{5} A_{6}$
$\frac{2}{6 _{C_{3}}} = \frac{2}{20} = \frac{1}{10}$

Q. 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 :

103​

101​

203​

51​

Only two equilateral tringles are possible A1​A3​A5​ and A2​A5​A6​ 6C3​​2​=202​=101​

$\frac{3}{10}$

$\frac{1}{10}$

$\frac{3}{20}$

$\frac{1}{5}$

Only two equilateral tringles are possible $A_{1} A_{3} A_{5}$ and $A_{2} A_{5} A_{6}$
$\frac{2}{6 _{C_{3}}} = \frac{2}{20} = \frac{1}{10}$