Probability Ques 6
- Three of the six vertices of a regular hexagon are chosen at rondom. The probability that the triangle with three vertices is equilateral, equals
(1995, 2M)
(a) $1 / 2$
(b) $1 / 5$
(c) $1 / 10$
(d) $1 / 20$
Show Answer
Answer:
Correct Answer: 6.(c)
Solution: (c) 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 $\triangle A E C$ and $\triangle B F D$.
$\therefore \quad $ Favourable ways $=2$
So, required probability
$ =\frac{2}{20}=\frac{1}{10} $
BETA
