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

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) 1/2 (B) 1/5 (C) 1/10 (D) 1/20

Tardigrade Admin · Accepted Answer

Three vertices out of 6 can be chosen in 6C3 ways So, total ways =6C3 =20 Only two equilateral triangles can be formed Δ AEC and Δ BFD <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/4104f010845a5e4a785f41a9e0ceffbc-.png" /> ∴ Favourable ways = 2 So, required probability = (2/20)=(1/10)

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

1/2

1/5

1/10

1/20

Three vertices out of $6$ can be chosen in $^{6} C_{3}$ ways
So, total ways $=^{6} C_{3} = 20$
Only two equilateral triangles can be formed
$Δ A EC$ and $Δ BF D$

$∴$ Favourable ways $= 2$
So, required probability = $\frac{2}{20} = \frac{1}{10}$

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

1/2

1/5

1/10

1/20

Three vertices out of 6 can be chosen in 6C3​ ways So, total ways =6C3​=20 Only two equilateral triangles can be formed ΔAEC and ΔBFD ∴ Favourable ways =2 So, required probability = 202​=101​

Three vertices out of $6$ can be chosen in $^{6} C_{3}$ ways
So, total ways $=^{6} C_{3} = 20$
Only two equilateral triangles can be formed
$Δ A EC$ and $Δ BF D$

$∴$ Favourable ways $= 2$
So, required probability = $\frac{2}{20} = \frac{1}{10}$