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If three vertices of a regular hexagon are chosen at random, then the chance that they form an equilateral triangle is:
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Verified Answer
The correct answer is:
$\frac{1}{10}$
Three vertices can be selected in ${ }^{6} \mathrm{C}_{3}$ ways.
The only equilateral triangles possible are $\mathrm{A}_{1} \mathrm{~A}_{3} \mathrm{~A}_{5}$ and $\mathrm{A}_{2} \mathrm{~A}_{4} \mathrm{~A}_{6}$
$\mathrm{p}=\frac{2}{{ }^{6} \mathrm{C}_{3}}=\frac{2}{20}=\frac{1}{10}$
The only equilateral triangles possible are $\mathrm{A}_{1} \mathrm{~A}_{3} \mathrm{~A}_{5}$ and $\mathrm{A}_{2} \mathrm{~A}_{4} \mathrm{~A}_{6}$
$\mathrm{p}=\frac{2}{{ }^{6} \mathrm{C}_{3}}=\frac{2}{20}=\frac{1}{10}$
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