5. In a cube, the probability that three randomly chosen edges are pairwise skew is $\qquad$

In a cube, randomly select three edges, the probability that they are pairwise skew is $\qquad$
5. $\frac{2}{55}$.

Let the cube be $A B C D-E F G H$, with a total of 12 edges. The number of ways to choose any three edges from these is $\mathrm{C}_{12}^{3}=220$.
Next, consider the number of ways to choose three edges such that they are pairwise skew.
Since the edges of the cube determine three mutually non-parallel directions (i.e., the directions of $A B$, $A D$, and $A E$), edges with the same direction are pairwise coplanar. Therefore, the three chosen edges must belong to three different directions. We can first choose an edge in the $A B$ direction, which has four possible choices.

Assume the chosen edge is $A B$. Then, in the $A D$ direction, the edge can only be $E H$ or $F G$, giving two possibilities. When the edge in the $A D$ direction is $E H$ or $F G$, the edge in the $A E$ direction can only be $C G$ or $D H$, respectively.
In summary, the number of ways to choose three edges that are pairwise skew is 8.
Thus, the required probability is $\frac{8}{220}=\frac{2}{55}$.