A) \[\frac{1}{2}\]
B) \[\frac{1}{5}\]
C) \[\frac{1}{10}\]
D) \[\frac{1}{20}\]
Correct Answer: C
Solution :
Total number of triangles which can be formed is equal to \[{}^{6}{{C}_{3}}=\frac{6\times 5\times 4}{1\times 2\times 3}=20\] Number of equilateral triangles = 2 \[\therefore \] Required probability \[=\frac{2}{20}=\frac{1}{10}.\]
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