(2) 30 points on a circle divide it equally. If any 3 different points are taken, the probability that these 3 points, when connected in sequence, form an equilateral triangle is $\qquad$ .
(2) $\frac{1}{406}$ Hint: The number of ways to choose 3 different points from 30 points is $\mathrm{C}_{30}^{3}$, and there are 10 ways to form

an equilateral triangle, so the probability is $\frac{10}{\mathrm{C}_{30}^{3}}$.