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 distances between each pair of these three points do not exceed 2 is $\qquad$
As shown in the figure, the number of three-point groups corresponding to six edges is 6; the number of three-point groups corresponding to four small squares is $4 C_{4}^{3}=16$; the number of three-point groups corresponding to the vertices of the four sides and the center is 4; the number of three-point groups corresponding to the midpoints of the four sides forming a square is also 4.
Therefore, the total number of three-point groups that meet the conditions is $6+16+4+4=30$.
Let $A=$ “the distance between any two of the three points does not exceed 2”, so we have $P(A)=\frac{30^{\prime}}{C_{9}^{3}}=\frac{30}{84}=\frac{5}{14}$.