# Changing events when given additional information in Bayes' Theorem

I encountered a problem as follows:

You have a white cube with side length 3. You paint all the faces red, then cut it into 27 small cubes with side length 1. Then, you randomly pick one small cube, roll it, and see 5 white faces. What is the probability that the last face is red?

My initial thought was to apply Bayes' Theorem as so:

P(last face is red|see 5 white faces) = P(see 5 white faces | last face is red) $$\cdot$$ P(last face is red) / P(see 5 white faces)

The solution, however, does not use "last face is red" as the first event. Instead, it combines the information from "last face is red" and "given 5 white faces" to make the event "1 red face and 5 white faces", which occurs when the mini cube is a center cube (center of one of the sides).

Thus, the solution's application of Bayes' Theorem looks like

P(center cube | see 5 white faces) = P(see 5 white faces | center cube) $$\cdot$$ P(center cube) / P(see 5 white faces).

This is a lot easier to solve than the first approach - to get the probability of purely the "last face is red", I think you would need to use the law of total probability across the different types of mini cubes, and I'm not sure if that would even give the correct answer. On the other hand, finding the probability of picking a center cube as well as the probability of seeing 5 white sides given you picked a center cube are quite straightforward.

What's the difference between the first and second approach? When you are given additional information, are you supposed to "change" the event to another that includes the given information?

• Was this problem from a probability book , or just a brain teaser ? Commented Nov 6, 2023 at 7:00
• I found the problem online. Commented Nov 6, 2023 at 7:07

This is not changing the events, rather just changing the description of the events.

There are four types of minicubes. The true-centre, face-centre, edge, and corner. Of which there are one, six, twelve, and eight respectively, and which have zero, one, two, and three red faces respectively.

So:

$$\Pr(\textbf{last face is red}\mid\text{see 5 white faces})\\=\dfrac{\Pr(\textbf{last face is red }\&\text{ see 5 white faces})}{\Pr(\text{see 5 white faces})}\\=\dfrac{\Pr(\text{face-centre cube }\&\textbf{ lands red face down})}{\Pr(\text{see 5 white faces})}\\=\dfrac{\Pr(\text{face-centre cube })\Pr(\textbf{ lands red face down}\mid\text{face-centre cube})}{\Pr(\text{see 5 white faces})}$$

Well, either way, you use the Law of Total Probability to calculate the probability for seeing five white faces. Since you shall see five white faces if the cube is either: the true-centre cube, or it is a face-centre cube that lands red face down.

$$\Pr(\text{see 5 white faces})={\Pr(\text{face-centre cube }\&\textbf{ lands red face down})+\Pr(\text{true-centre cube})}$$

• It seems like this is what's happening: Let A = last face is red, and B = see 5 white faces. Instead of writing the problem as P(A|B), reword it as P(A and B | B). "A and B" is equivalent to "face-center cube and the red face is down". Now we get the concept of "face-center cube", and it becomes much easier to solve. Commented Nov 12, 2023 at 8:03