5. As shown in Figure 3, in a $4 \times 4$ dot array, the probability of forming a triangle by randomly selecting three points is ( ).
(A) $\frac{129}{140}$
(B) $\frac{67}{70}$
(C) $\frac{27}{28}$
(D) $\frac{41}{42}$
5. A.

The number of ways to choose any three points (disregarding order) is $\mathrm{C}_{16}^{3}=560$, and the number of ways that do not form a triangle (i.e., the three points are collinear) is 44. Therefore,
$$
P=\frac{560-44}{560}=\frac{129}{140} .
$$