Probability 1 Question 3
3. If there of the six vertices of a regular hexagon are chosen at random, then the probability that the triangle formed with these chosen vertices is equilateral is
(a) $\frac{1}{10}$
(b) $\frac{1}{5}$
(c) $\frac{3}{10}$
(d) $\frac{3}{20}$
(2019 Main, 12 April I)
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Answer:
Correct Answer: 3. (a)
Solution:
- Since, there is a regular hexagon, then the number of ways of choosing three vertices is ${ }^{6} C _3$. And, there is only two ways i.e. choosing vertices of a regular hexagon alternate, here $A _1, A _3, A _5$ or $A _2, A _4, A _6$ will result in an equilateral triangle.
$\therefore$ Required probability
$$ =\frac{2}{{ }^{6} C _3}=\frac{2}{\frac{6 !}{3 ! 3 !}}=\frac{2 \times 3 \times 2 \times 3 \times 2}{6 \times 5 \times 4 \times 3 \times 2 \times 1}=\frac{1}{10} $$
