MHT CET · Maths · Probability
Three of six vertices of a regular hexagon are chosen at random. The probability that the triangle with these three vertices is equilateral, equals
- 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
The number of triangles that can be drawn using 6 vertices is given by
\(\mathrm{n}(\mathrm{S})={ }^6 \mathrm{C}_3=20\)
\(A\) : Event of selecting equilateral triangle. The equilateral triangle can be drawn if selected three vertices are alternate.
\(\begin{aligned} & \therefore\mathrm{n}(\mathrm{A})=2 \\ & \therefore \mathrm{P}(\mathrm{A})=\frac{2}{20}=\frac{1}{10}\end{aligned}\)
\(\mathrm{n}(\mathrm{S})={ }^6 \mathrm{C}_3=20\)
\(A\) : Event of selecting equilateral triangle. The equilateral triangle can be drawn if selected three vertices are alternate.
\(\begin{aligned} & \therefore\mathrm{n}(\mathrm{A})=2 \\ & \therefore \mathrm{P}(\mathrm{A})=\frac{2}{20}=\frac{1}{10}\end{aligned}\)
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