Three of the six vertices of a regular hexagon are chosen at random. T

1.	Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is equilateral equals :(a) \(\frac{1}{6}\)(b) \(\frac{1}{3}\)(c) \(\frac{1}{10}\)(d) None of these
Answer» (c) \(\frac{1}{10}\) Let S be the sample space. Then n(S) = Number of triangles formed by selecting any three vertices of 6 vertices of a regular hexagon =⁶C₃= \(\frac{6\times5\times4}{3\times2}\) = 20. Let A : Event that the selected three vertices form an equilateral triangle. Then n(A) = 2 (As only two equilateral triangles are formed from the vertices of a regular hexagon) ∴ Required probability = \(\frac{n(A)}{n(S)}\) = \(\frac{2}{20}\) = \(\frac{1}{10}\).

Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is equilateral equals :(a) \(\frac{1}{6}\)(b) \(\frac{1}{3}\)(c) \(\frac{1}{10}\)(d) None of these

Answer»

(c) \(\frac{1}{10}\)

Let S be the sample space.

Then n(S) = Number of triangles formed by selecting any three vertices of 6 vertices of a regular hexagon

=⁶C₃= \(\frac{6\times5\times4}{3\times2}\) = 20.

Let A : Event that the selected three vertices form an equilateral triangle.

Then n(A) = 2

(As only two equilateral triangles are formed from the vertices of a regular hexagon)

∴ Required probability = \(\frac{n(A)}{n(S)}\) = \(\frac{2}{20}\) = \(\frac{1}{10}\).

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Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is equilateral equals :(a) \(\frac{1}{6}\)(b) \(\frac{1}{3}\)(c) \(\frac{1}{10}\)(d) None of these

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