The most transparent exposition of Bayes' Theorem

I am seeking the most transparent exposition of Bayes' Theorem (for undergraduates). I would prefer to avoid mentioning "prior" and "posterior," and instead focus on frequencies. The Wikipedia entry is quite good in some ways, but I disagree with this:

"The role of Bayes’ theorem is best visualized with tree diagrams."

The tree diagrams do little for me. But the associated table is quite useful:

Wikipedia image.

So $$P( A \& B ) = \frac{2}{20}$$.

$$P( A | B ) = \frac{2}{8}$$ and $$P(B) = \frac{8}{20}$$. So $$P( A \& B ) = P( A | B )\; P(B) = \tfrac{2}{8} \tfrac{8}{20} = \tfrac{2}{20}.$$

$$P( B | A ) = \frac{2}{5}$$ and $$P(A) = \frac{5}{20}$$. So $$P( A \& B ) = P( B | A )\; P(A) = \tfrac{2}{5} \tfrac{5}{20} = \tfrac{2}{20}.$$

And this leads to $$P( A \& B ) = P( A | B )\; P(B) = P( B | A )\; P(A),$$ which is one form of Bayes' Theorem.

So here are my questions:

Q1. What is a compelling story one can associate with this specific table, identifying $$A$$ and $$B$$ with real-world events that make sense with that table's data?

Q2. What is the clearest justification one could make for $$P( A \& B ) = P( A | B )\; P(B)$$, independent of any particular interpretation of $$A$$ and $$B$$?

• Can you just have two groups of students who are sitting at Table A and Table B? Say, e.g., 2 wearing red and 3 wearing blue at Table A, and 6 wearing red and 9 wearing blue at Table B. I think students can derive the identities for Bayes' Theorem from such a setup. Aug 25 '19 at 3:07
• +: If tree diagrams don't do it for you, then maybe Venn Diagrams might? Aug 25 '19 at 3:09
• Isn’t the Wikipedia image above already basically a Venn diagram, albeit with rectangles instead of the usual circles?
– Joe
Aug 26 '19 at 12:34
• I think the table makes things more complicated. It has more structure and information than is necessary. It would be better to have a circle that is the whole event space, with a portion shaded representing A, and then a circle within the outer circle representing B. You can infer Pr[A|B] is the portion of B that is shaded, which is Pr[A∧B]/Pr[B]. Aug 26 '19 at 17:58
• 3Blue1Brown has a version of Bayes' rule which has a simpler formula: youtube.com/watch?v=lG4VkPoG3ko Apr 2 '21 at 16:04

Q2. One way I think about this is to view the conditional probability as being defined by $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$
and $$P(A|B) := 0$$ when $$P(B) = 0$$ rather than defining the probability of the intersection as the product of the other two probabilities.
The exposition that goes along with it is "Assume within our universe, event B has occurred (i.e., given $$B$$). Then $$P(A|B)$$ is the probability of $$A$$ in the new, restricted universe".