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14.Probability
hard
Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these three vertices is equilateral, is equal to
A
$\frac{1}{2}$
B
$\frac{1}{5}$
C
$\frac{1}{{10}}$
D
$\frac{1}{{20}}$
(IIT-1995)
Solution
(c) 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}}.$
Standard 11
Mathematics
