# Thread: Need help with a probability and a cube question.

1. ## Need help with a probability and a cube question.

A 4"x4"x4" cube is painted and then cut into sixty-four 1"x1"x1" cubes. A unit cube is then randomly selected and rolled. What is the probability that the top face of the rolled cube is painted? Express your answer as a common fraction.

Now, originally, I tohught the answer was 96/384, but I realized it was wrong since some sides (the corner cubes) have more than one painted side.

2. Originally Posted by help1
A 4"x4"x4" cube is painted and then cut into sixty-four 1"x1"x1" cubes. A unit cube is then randomly selected and rolled. What is the probability that the top face of the rolled cube is painted? Express your answer as a common fraction.
Now, originally, I tohught the answer was 96/384, but I realized it was wrong since some sides (the corner cubes) have more than one painted side.
There are cubes with no painted sides.
There are cubes with one painted side.
There are cubes with two painted sides.
There are cubes with three painted sides.
How many of each?

3. This is actually a cool problem. There are patterns to the number of painted faces on a nXn cube.

For a 3X3X3 cube, there are 8 cubes with 3 painted faces. Any cube will always have 8 cubes with 3 painted faces. Those are the corners.

There are 12 cubes with 2 faces painted. 6 cubes with 1 face painted and 1 with no faces painted(the one in the dead center).

That is 54 painted faces with 162 possible sides. 1/3 probability.

Now, for the 4X4X4 cube:

There are 8 with 3 faces painted. 24 with 2 faces painted. 24 with 1 face painted. 8 with 0 faces painted. That's 96 painted faces out of 384 possible sides. 96/384=1/4 probability. You are correct.

Now for a 5X5X5:

There are 8 with 3 faces painted. 36 with 2 painted. 54 with 1 painted and 27 with 0 painted

That's 150 painted sides out of 750 sides. 150/750=1/5.

See the pattern?. The probability of rolling a painted side is 1/n.

If it's a 10X10X10 cube, it's 1/10. Cool, huh?.

Therefore, we have a formula for the number of 2-painted sides: 12(n-2)

The number of 3 painted sides is always 8.

The number of 1-painted sides: $6(n-2)^{2}$

The number of no painted sides: $(n-2)^{3}$

The number of cubes is then $(n-2)^{3}+6(n-2)^{2}+12(n-2)+8=n^{3}$