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 there exist two points among these three with a distance of $\sqrt{5}$ is $\qquad$
As shown in the figure, there are 8 pairs of points with a distance of $\sqrt{5}$: $14,18,27,29$,
$36,38,47,69$. Each pair of points corresponds to 7 triplets of points, and each 2 pairs of points with a common point result in one repeated count of triplets: $148,147,138,279,247,269,368,369$.
Therefore, the number of triplets that satisfy the condition is $7 \times 8-8=48$. Let $A=$ “among the three points, there exist two points with a distance of $\sqrt{5}$ ”, hence $P(A)=\frac{48}{C_{9}^{3}}=\frac{48}{84}=\frac{4}{7}$.