5. Choose three vertices from the eight vertices of a cube. Then the probability that these three points form a right-angled triangle is
5. $\frac{6}{7}$.

Solution 1: From the eight vertices of a cube, three can be chosen, with a total of $\mathrm{C}_{8}^{3}=56$ ways to choose.

A cube has six faces and six diagonal planes, all of which are rectangles (including squares). In each rectangle, there are four right-angled triangles, so there are a total of $12 \times 4=48$ right-angled triangles.
Therefore, the probability of forming a right-angled triangle is $P=\frac{48}{56}=\frac{6}{7}$.
Solution 2: From the eight vertices of a cube, three can be chosen, with a total of $\mathrm{C}_{8}^{3}=56$ ways to choose.

First, find the number of acute triangles (i.e., equilateral triangles). It is easy to see that each vertex corresponds to a unique equilateral triangle. Therefore, there are 8 acute triangles.

No three points can form an obtuse triangle. Thus, the number of right-angled triangles is $56-8=48$.
Therefore, the probability of forming a right-angled triangle is $P=\frac{48}{56}=\frac{6}{7}$.