2005 AIME I Problem 9

Each small cube’s two unpainted faces share one edge of that cube; the surface is all orange exactly when every such edge is buried

A random orientation puts that edge uniformly among the $12$ edge positions. A corner cube hides $3$ of them, an edge cube $5,$ a face cube $8.$

Each unit cube has one “bad edge”: the edge shared by its two unpainted faces. The larger cube’s surface is entirely orange exactly when every unit cube’s bad edge touches no visible face. A uniformly random orientation places the bad edge uniformly among the cube’s $12$ edge positions, so for each unit cube we count the edge positions both of whose faces are hidden.

A corner cube shows $3$ faces meeting at a vertex; the safe edges are those of the $3$ hidden faces meeting at the opposite vertex, so the probability is $\frac{3}{12} = \frac{1}{4}.$ An edge cube shows $2$ adjacent faces, which touch $4 + 4 - 1 = 7$ edges, leaving $5$ safe: probability $\frac{5}{12}.$ A face-center cube shows $1$ face touching $4$ edges, leaving $8$ safe: probability $\frac{8}{12} = \frac{2}{3}.$ The center cube is always fine.

With $8$ corner, $12$ edge, and $6$ face-center cubes, the probability is $$\begin{aligned} &\left(\frac{1}{4}\right)^{8}\left(\frac{5}{12}\right)^{12}\left(\frac{2}{3}\right)^{6} \\ &= \frac{5^{12}}{2^{34} \cdot 3^{18}}, \end{aligned}$$ so $$\begin{aligned} &a + b + c + p + q + r \\ &\quad {}= 12 + 34 + 18 \\ &\qquad {}+ 5 + 2 + 3 \\ &\quad {}= 74. \end{aligned}$$