A cube 4 units on each side is composed of 64 unit cubes. Two faces of the larger cube that share an edge are painted blue, and the cube is disassembled into 64 unit cubes. Two of the unit cubes are selected uniformly at random. What is the probability that one of two selected unit cubes will have two painted faces while the other unit cube has no painted faces?
+129850 Here's my best attempt
We will have 2(3*4) = 24 cubes with one painted face and 4 cubes with 2 painted faces
Note that we could either choose a cube with two painted faces first, or one with one painted face first....so....we have two ways of doing this....so..we have
P( two painted faces and one painted face) = 2 P(two painted faces)(one painted face) =
2 (4/64) (24/64) = (1/8)(24/64) = 3 /64
