AP EAMCET · Maths · Probability
3 out of 6 vertices of a regular hexagon are chosen at a time at random. The probability that the triangle formedwith these three vertices is an equilateral triangle, is
- A \(\frac{1}{2}\)
- B \(\frac{1}{5}\)
- C \(\frac{1}{10}\)
- D \(\frac{1}{20}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{10}\)
Step-by-step Solution
Detailed explanation
Give that, 3 out of 6 vertices of hexagon are chosen. Total number of outcomes \(={ }^5 \mathrm{C}_3=\frac{6 !}{3 ! 3 !}=20\) Only two equilateral triangles are possible of regular hexagon, i.e. \(\triangle \mathrm{AEC}\) and \(\triangle \mathrm{BFD}\).…
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