# Painted Cube and the Law of Total Probability

Problem:

Suppose you are given a white cube that is broken into $$27$$ pieces. This cube was originally $$3$$ units by $$3$$ units, and has been broken into $$27$$ smaller cubes, witch each smaller cube having dimensions $$1 \times 1 \times 1$$. Before this cube was broken, all $$6$$ of its faces were painted green. You randomly pick a small cube and see that $$5$$ faces are white. What is the probability that the bottom face is also white?

In this conditional probability problem, I am having trouble determining $$P(A)$$ and $$P(A^c)$$, which are critical to determining $$P(B)$$ via the Law of Total Probability.

$$P(B|A)$$ is relatively straightforward, since it is the probability of picking a mini-cube with 5 White Faces given that the Bottom Face is White. Only 1 mini-cube satisfies that, the Core Cube, so it's probability ought to be $$1/27$$.

Likewise, $$P(B|A^c)$$ is pretty easy, its the probability of 5 White Faces given that the Bottom Face is Not White (Green). Only 6 mini-cubes satisfy that, the ones in the centers of the 6 faces of the original cube. This makes the probability $$6/27$$.

But what of $$P(A)$$, the probability of "Bottom White"? Originally I thought it was $$1/27$$, since there is only 1 cube that meets that configuration, but upon checking my answer it seems $$P(A) = 1$$. $$P(A^c)$$ is the probability that the Bottom is not White, so originally I went with $$6/27$$, the number of mini-cubes that could have 5 white faces and still have a bottom face that is not white. However, the solution maintains $$P(A^c) = 1/6$$.

What is the reasoning that leads to $$P(A) = 1$$ and $$P(A^c) = 1/6$$?

• The probability of an event and it's complement must add to 1. So perhaps what they call A and A complement do not represent Bottom white and Bottom Not White. Commented May 3, 2023 at 1:09
• There is a crucial unanswered question. Are we to assume that each face of the smaller cubes is equally likely to be the face that you cannot see? If that's the case, then the conditional probability that the bottom face is also white is $\frac 12$. Commented May 3, 2023 at 1:51
• Yes, that sounds right. Commented May 3, 2023 at 15:48

I'm going to assume (a) each small cube is equally likely to be chosen, and (b) each face of the cube you chose is equally likely to be the bottom face.

If these two assumptions are correct, then given that you can see $$5$$ white faces, there could be any $$12$$ equally likely possible faces on the bottom. Of those, $$6$$ are white (the $$6$$ faces of the central cube) and $$6$$ are green (the outer face of the $$6$$ cubes at the center of each face of the larger cube). Therefore, given that you can see $$5$$ white faces, the probability that the bottom face of your chosen cube also is white is $$\frac 12$$.

• I was expecting a solution using Bayes Theorem, but this is a really good solution. Its very intuitive. Commented May 3, 2023 at 15:58
• I am confused by the answer: why the center white cube count for 6 configurations? It is intuitive for me that if i see five white faces, then i only have 7 options (6side cubes and 1 center cube)
– Rain
Commented Aug 12 at 16:53
• I think I understand it now: really tricky with the question wording: giving five faces white, the center cube can indeed account for six configurations due to any side of it lying on the bottom is a possibility.
– Rain
Commented Aug 12 at 17:26

Let $$A$$ the event "the bottom face is white" and $$B$$ the event "the 5 faces you see are white". Moreover, for $$i\in \{5,6\}$$, let $$C_i$$ the event "the selected cube has exactly $$i$$ white faces".

Assume each face of the selected cube is equally likely to be the bottom face, so that $$P(B|C_5) = \frac 16$$.

We look for \begin{align} P(A|B) = \frac{P(A\cap B)}{P(B)} &= \frac{P(C_6)}{P(B|C_6)P(C_6) + P(B|C_5)P(C_5) + P(B\cap (C_5\cup C_6)^c)} \\&= \frac{1/27}{1\times \frac 1{27} + \frac 1{6}\times \frac 6{27} + 0} \\&= \frac 12. \end{align}