imc

First of all, number of planes determined by any three vertices of a cube is $20$ ($6$ surface, $6$ opposing parallel edges, $8$ points cut by three remote vertices). Among these $20$ planes only $6$ surfaces will not cut into the cube. Secondly, to choose three vertices randomly, the four vertices planes each will be chosen $4$ times, while the three vertices planes each will be chosen once. To conclude, the probability of a cutting in plane is $\frac{(6\cdot 4+8\cdot 1)}{(12\cdot 4+8\cdot 1)}=\frac{32}{56}$ = $\boxed{\frac{4}{7}}$.