Three colours meeting at a corner of a cube

The six faces of a cube are painted with six different colours, one of which is red, one blue and one green, with every assignment equally likely. What is the probability that the red, blue and green faces all meet at a single corner?

$2/5$. Three faces share a corner exactly when no two of them are opposite. Given red's face, blue avoids its opposite with probability $4/5$; then green must avoid the faces opposite both red and blue, leaving $2$ good faces of the $4$ remaining, so $\tfrac45\cdot\tfrac12=\tfrac25$.