Question

Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertics is not equilateral is

• A. $1/2$
• B. $1/5$
• C. $9/10$
• D. $1/20$

Answer 1

Answer: (C)

$9/10$

Three vertices out of $6(A,B,C,D,E,F)$ can be chosen in ${}^6{C}_3$ ways.
So, total ways${=}^6{C}_3=20$
only two equilateral triangles can be formed $△AEC$ and $△BFD$.
So, favourable ways $=2$
Probability$=\frac{2}{20}=\frac{1}{10}$
Hence, required probability$=1-\frac{1}{N}=\frac{9}{10}$