# Question

In this exercise we are introduced to the intersection of events A and B when they're independent. How does this change when A and B are dependent?# Answer

A helpful idea is to look at what happens to Bayes' theorem when A and B are independent. First, note that because A and B do not depend on one another, P(A|B) must equal P(A) and likewise P(B|A) = P(B). With that note, let's look at Bayes' theoremP(A|B) = P(B|A) * P(A)/P(B)

Substituting P(B) for P(B|A) gives us

P(A|B) = P(B) * P(A) / P(B)

We could immediately divide the P(B) in the numerator and denominator but it is better to first acknowledge that the numerator P(A) * P(B) is exactly what P(A ∩ B) equals. This is very convenient but it’s not a coincidence. The numerator of Bayes’ theorem is another way of writing P(A ∩ B).

So to think about P(A ∩ B) in the case where A and B are dependent, use the numerator of Bayes’ theorem.