Consider this regular triangular lattice:

If three dots are chosen at random, what is the probability that they will form the vertices of an equilateral triangle?

If the probability is given by $\dfrac{a}{b}$ where $a$ and $b$ are coprime positive integers, express your answer as $a+b$ .

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The answer is 9.

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There are 15 ways to get an equilateral triangle:

And, there are $\binom{10}{3} = 120$ ways to choose 3 dots.

Therefore, the probability is $\dfrac{15}{120} = \dfrac{1}{8}$

$1+8 = \boxed{9}$