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?

• Your computation is wrong in a number of ways. First of all, there are $6^3$ unpainted small cubes. Secondly, all you do is to compute the probability that a randomly selected cube us unpainted...you ignore the fact that the roll came up (or rather, down) unpainted.
– lulu
Commented May 19, 2019 at 19:38

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:

$$P(unpainted|blank)=\frac{P(unpainted\ and\ blank)}{P(blank)}=\frac{P(unpainted)}{P(blank)}=\frac{216/512}{7/8}=\frac{27}{56}\approx 0.48$$

• Thanks for your reply! I will use your hint and further try to solve the problem. Commented May 19, 2019 at 19:36
• Typo: you said $6^6 = 216$ Commented May 19, 2019 at 21:19
• @NeilA. Thanks. Corrected. Commented May 19, 2019 at 21:22

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.