+355 The six faces of a cube are painted black. The cube is then cut into 6^3 smaller cubes, all the same size. One of the small cubes is chosen at random, and rolled. What is the probability that when it lands, the face on the top is black?
+346 To solve this problem, let's first consider the structure of the cube.
When the large cube is cut into \(6^3\) smaller cubes, each small cube will have one face painted black and the remaining faces unpainted (since each face of the large cube was painted black initially).
Now, if we roll one of these small cubes, regardless of which face is on top initially, it has a \(1/6\) chance of landing with each face facing upward, since all faces are identical.
However, the probability that the face on top after rolling is black depends on whether the cube was rolled along an axis perpendicular to a black face or not. If it was, the black face will still be on top; if not, it won't be.
Considering that the black faces are on the outside of the original cube, any roll along the x, y, or z-axis will result in the black face being on top. There are 3 such axes.
So, the probability of the black face being on top after rolling is:
\[P(\text{Black face on top}) = \frac{3}{6} = \frac{1}{2}\]
Hence, the probability that when the small cube is rolled, the face on the top is black is \( \frac{1}{2} \).
