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



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Answer: $\frac{2}{55}$. Solution: Let the cube be ABCD-EFGH, it has a total of 12 edges, and the number of ways to choose any 3 edges from them is $C_{12}^{3}=220$.
Now consider the number of ways to select 3 edges that are pairwise skew. Since the edges of the cube determine 3 mutually non-parallel directions (i.e., the directions of $\mathrm{AB}$, $\mathrm{AD}$, and $\mathrm{AE}$), the 4 edges with the same direction are pairwise coplanar, so the 3 selected edges must belong to 3 different directions. We can first choose an edge in the $\mathrm{AB}$ direction, which has 4 possible choices. Suppose the chosen edge is $\mathrm{AB}$, then in the $\mathrm{AD}$ direction, only the edge $\mathrm{EH}$ or the edge $\mathrm{FG}$ can be chosen, for a total of 2 possibilities. When the edge chosen in the $\mathrm{AD}$ direction is $\mathrm{EH}$ or $\mathrm{FG}$, the edge in the $\mathrm{AE}$ direction can only be CG or DH, respectively. From the above, the number of ways to select 3 edges that are pairwise skew is $4 \times 2=8$, so the required probability is $\frac{8}{220}=\frac{2}{55}$.