An 8x8 cube is painted and cut into 1x1 cubes. One of them is taken a rolled, with bottom being blank. Probability that entire 1x1 cube is unpainted? 
Andy has a cube of edge length 8 cm. He paints the outside of the cube red and then divides the cube into smaller cubes, each of edge length 1 cm. Andy randomly chooses one of the unit cubes and rolls it on a table. If the cube lands so that an unpainted face is on the bottom, touching the table, what is the probability that the entire cube is unpainted?

My attempt:
Probability=$\frac{\text{possible cubes}}{\text{total cubes}}$.
Since the bottom face is unpainted, and we want the visible faces unpainted too, then the entire cube must be unpainted. There are  $6\cdot 64=384$ of these cubes.
There are 512 cubes in total.
Thus, the probability must be $\frac{384}{512}=\frac{3}{4}$.
However, the correct answer is $\frac{27}{56}$. Where am I wrong?
 A: Hint
Bayes' theorem. It will be helpful to consider the $4$ kinds of cubes: from the corner of the initial $8\times8$ cube, or from an edge, or from the interior of a face, or from inside the cube. Count how many there are of each kind (and how many painted faces they have), then the probability to roll a blank bottom for each case. Then apply Bayes' theorem.

Detailed solution
(note that the number of unpainted cubes is not $6\times64=384$ but $6^3=216$)
Corner cubes: $8$ such cubes, with $3$ faces painted (hence probability $3/6=1/2$ to roll a blank bottom.
Edge cubes (and not corner): $6$ on each edge, and there are $12$ edges, hence $72$ such cubes, with $2$ painted faces each. Probability to roll a blank bottom: $4/6=2/3$.
Face cubes (and not corner nor edge): $6\times6=36$ on each face, and there are $6$ faces, hence $216$ such cubes, with $1$ painted face each. Probability to roll a blank bottom: $5/6$.
Inner cubes: $6^3=216$ such cubes, and they have no painted face. Probability to roll a blank bottom: $1$.
Quick check: $216+216+72+8=512=8^3$.
What is the probability to roll a blank bottom (called event "blank" below)?
$$P(blank)=P(corner)\cdot P(blank|corner)+P(edge)\cdot P(blank|edge)\\+P(face)\cdot P(blank|face)+P(inner)\cdot P(blank|inner)\\=\frac{8}{512}\cdot\frac{3}{6}+\frac{72}{512}\cdot\frac{4}{6}+\frac{216}{512}\cdot\frac{5}{6}+\frac{216}{512}\cdot\frac{6}{6}=\frac78$$
Probability to have an unpainted cube given that the bottom is blank:
$$P(unpainted|blank)=\frac{P(unpainted\ and\ blank)}{P(blank)}=\frac{P(unpainted)}{P(blank)}=\frac{216/512}{7/8}=\frac{27}{56}\approx 0.48$$
A: Assuming that for each cube, there is an equal probability $\frac{1}{6}$ for each of the six faces to come at the bottom after the roll.
Now, for a cube of length $8$ cm painted at the surface and then divided into cubes each of length $1$ cm, we have $8$ cubes with three sides painted, $72$ cubes with two sides painted, $216$ cubes with one side painted and $216$ cubes with no side painted at all.
For cubes with $3$ sides painted, the probability of landing with an unpainted face down is equal to $\frac{3}{6}$.
For cubes with $2$ sides painted, the probability of landing with an unpainted face down is equal to $\frac{4}{6}$.
For cubes with $1$ sides painted, the probability of landing with an unpainted face down is equal to $\frac{5}{6}$.
For cubes with no sides painted, the probability of landing with an unpainted face down is $1$.
If $A_i, \, i = 0,1,2,3$ denotes the event of choosing a cube with $i$ sides painted. And $B$ denoted the event of landing with an unpainted face at the bottom, we get from the Bayes' Theorem that,
$P(A_0 \mid B) = \displaystyle \frac{P(A_0)*P(B\mid A_0)}{P(A_0)*P(B\mid A_0)+P(A_1)*P(B\mid A_1)+P(A_2)*P(B\mid A_2)+P(A_3)*P(B\mid A_3)}$
You just need to put the numbers now.
