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

Question

Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with three vertices is equilateral, equals (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Three vertices out of 6 can be chosen in 6C3 ways So, total ways =6C3=20 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/9614e68a38b94650047211120a463860-.png" /> only two equilateral triangles can be formed Δ AEC and Δ BFD So, favourable ways = 2. ∴ required probability =(2/20)=(1/10)=0.1

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

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 $Δ BF D$
So, favourable ways = $2$ .
$∴$ required probability $= \frac{2}{20} = \frac{1}{10} = 0.1$

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

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 So, favourable ways = 2. ∴ required probability =202​=101​=0.1

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 $Δ BF D$
So, favourable ways = $2$ .
$∴$ required probability $= \frac{2}{20} = \frac{1}{10} = 0.1$