3. When three dice are thrown simultaneously, the probability that the sum of the numbers facing up is no more than 7 is ( ).
(A) $\frac{35}{6^{3}}$
(B) $\frac{34}{6^{3}}$
(C) $\frac{31}{6^{3}}$
(D) $\frac{30}{6^{3}}$
3.A.

Let the numbers rolled by the three dice be $x, y, z$, then the number of positive integer triples $(x, y, z)$ that satisfy $x+y+z \leqslant 7$ is
$$
C_{6}^{2}+C_{5}^{2}+C_{4}^{2}+C_{3}^{2}+C_{2}^{2}=35 \text { (sets) } \text {. }
$$

Therefore, the probability of the situation described in the problem is $\frac{35}{6^{3}}$.