2009 AMC 10A Problem 24

There are $\binom{8}{3} = 56$ equally likely triples of vertices, and any three vertices of a cube are non-collinear, so each triple determines a plane.

Count the complement: triples whose plane stays outside the interior. If the three vertices all lie on one face, their plane is that face's plane and touches only the boundary. Each of the $6$ faces has $4$ vertices, giving $\binom{4}{3} = 4$ triples per face, and no triple lies on two faces (two faces share at most $2$ vertices). That is $6 \cdot 4 = 24$ "bad" triples.

Conversely, if the three vertices are not all on one face, their plane enters the interior. Put the cube at $[0,1]^3$. The centroid of the three vertices lies in their plane, and each of its coordinates is $0$ or $1$ only if all three vertices share that coordinate, i.e. lie on a common face. Otherwise every coordinate is $\frac13$ or $\frac23$, so the centroid is strictly inside the cube.

So $56 - 24 = 32$ triples work, and the probability is $\frac{32}{56} = \frac{4}{7}$.

The answer is $\boxed{\textbf{(C)}\ \frac{4}{7}}$.

The "good" planes are messy (some are diagonal rectangles, some are equilateral or right triangles cutting corners), but the "bad" ones are exactly the six face planes, so counting the complement is far easier. The centroid argument is a clean way to certify that any non-face triple really does pierce the interior.

• Miscounting the coplanar triples on a face (there are 4 per face, not 1) or treating the 6 diagonal rectangles through the cube as planes that miss the interior.
• Counting distinct planes instead of triples of vertices; the probability is over the $56$ vertex triples, not over the planes they determine.