6. In the Cartesian coordinate system $x O y$, the point set $K=\{(x, y) \mid x, y=-1,0,1\}$. If three points are randomly selected from $K$, then the probability that the distance between each pair of these three points is no more than 2 is $\qquad$ .
Answer: $\frac{5}{14}$.
Solution: Note that $K$ contains a total of 9 points, so the number of ways to randomly select three points from $K$ is $\mathrm{C}_{9}^{3}=84$.
When the distance between any two of the three selected points does not exceed 2, there are the following three cases:
(1) The three points lie on the same horizontal or vertical line, with 6 cases.
(2) The three points are the vertices of an isosceles right triangle with side lengths of $1,1, \sqrt{2}$, with $4 \times 4=16$ cases.
(3) The three points are the vertices of an isosceles right triangle with side lengths of $\sqrt{2}, \sqrt{2}, 2$, where there are 4 cases with the right-angle vertex at $(0,0)$, and 1 case each with the right-angle vertex at $( \pm 1,0),(0, \pm 1)$, totaling 8 cases.

In summary, the number of cases where the distance between any two of the three selected points does not exceed 2 is $6+16+8=30$, and thus the required probability is $\frac{30}{84}=\frac{5}{14}$.