There are six vertices of a regular hexagon are chosen at random, then the possibility that the triangle with three vertices is equilateral, is equal to:

Question

There are six vertices of a regular hexagon are chosen at random, then the possibility that the triangle with three vertices is equilateral, is equal to: (A)  (1/2)  (B)  (1/3)  (C)  (1/10)  (D)  (1/20)

Tardigrade Admin · Accepted Answer

There are six equilateral triangle is formed in the regular hexagon.
The total number of triangle formed by the six vertices

=^{6} C_{3} = 20

.
The total number of equilateral triangles formed

= 6

∴

Required probability

= \frac{2}{20}

= \frac{1}{10}

Q. There are six vertices of a regular hexagon are chosen at random, then the possibility that the triangle with three vertices is equilateral, is equal to:

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{1}{10}$

$\frac{1}{20}$

There are six equilateral triangle is formed in the regular hexagon.
The total number of triangle formed by the six vertices $=^{6} C_{3} = 20$ .
The total number of equilateral triangles formed $= 6$
$∴$ Required probability $= \frac{2}{20}$
$= \frac{1}{10}$

Q. There are six vertices of a regular hexagon are chosen at random, then the possibility that the triangle with three vertices is equilateral, is equal to:

21​

31​

101​

201​

There are six equilateral triangle is formed in the regular hexagon. The total number of triangle formed by the six vertices =6C3​=20. The total number of equilateral triangles formed =6 ∴ Required probability =202​ =101​

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{1}{10}$

$\frac{1}{20}$

There are six equilateral triangle is formed in the regular hexagon.
The total number of triangle formed by the six vertices $=^{6} C_{3} = 20$ .
The total number of equilateral triangles formed $= 6$
$∴$ Required probability $= \frac{2}{20}$
$= \frac{1}{10}$