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# Need clarity, kindly explain! iIf three 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 :

iIf three 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 :

• Option 1)

$\frac{1}{5}$

• Option 2)

$\frac{3}{20}$

• Option 3)

$\frac{3}{10}$

• Option 4)

$\frac{1}{10}$

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Only two equilateral triangle are possible $A_{1}A_{3}A_{5}$ and $A_{21}A_{4}A_{6}$

$\frac{2}{6C_{_{_{5}}}}=\frac{2}{20}=\frac{1}{10}$

Option 1)

$\frac{1}{5}$

Option 2)

$\frac{3}{20}$

Option 3)

$\frac{3}{10}$

Option 4)

$\frac{1}{10}$

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