The six faces of a cube are painted black. The cube is then cut into $3^3$ smaller cubes, all the same size.
(a) How many of the smaller cubes have exactly one black face?
(b) How many of the smaller cubes do not have any black faces?
(c) 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?
(a) There are three cases to consider: the black face can be on a face of the big cube, on an edge of the big cube, or on a corner of the big cube.
If the black face is on a face of the big cube, there are 6 faces, so there are $6 \cdot 1^2 = 6$ small cubes with exactly one black face in this case.
If the black face is on an edge of the big cube, there are 12 edges, so there are $12 \cdot 2^2 = 48$ small cubes with exactly one black face in this case.
If the black face is on a corner of the big cube, there are 8 corners, so there are $8 \cdot 3^2 = 72$ small cubes with exactly one black face in this case.
Thus, the total number of small cubes with exactly one black face is $6 + 48 + 72 = \boxed{126}$.
(b) There are $3^3 = 27$ small cubes, and we have found that there are $126$ small cubes with exactly one black face. Therefore, there are $27 - 126 = \boxed{-99}$ small cubes that do not have any black faces.
(c) Since there are 6 black faces on the big cube and the cube is cut into $3^3 = 27$ small cubes, there is a total of $6 \cdot 27 = 162$ black faces on the small cubes. The probability that the face on top of a randomly chosen small cube is black is equal to the probability that we pick a black face out of the 162 black faces at random. Therefore, the probability is $\boxed{\frac{162}{162} = 1}$.