## Question

Bayes’ theorem provides a way of computing `P(A|B)`

but uses `P(B|A)`

. How would we go about finding either without one being given initially?

## Answer

To think about this question more clearly, we need to think about the intersection: P(A ∩ B). Recall that P(A ∩ B) is the probability that both `A`

and `B`

are true. We learned about P(A ∩ B) when `A`

and `B`

were independent but it also makes sense to think about the intersection in other cases.

First, let’s rephrase the conditional probability in plain language. `P(A|B)`

is asking:

What is the probability that

`A`

will happen if we already know that`B`

happened?

Written like this makes it clear that we’re asking a question about both `A`

and `B`

happening. So `P(A|B)`

is related to P(A ∩ B) in some way but how exactly? There certainly not equal. This is where our knowledge that `B`

already happened comes into play. Since `B`

happened, the probability that we’re computing is essentially no longer between `0`

and `1`

but instead between `0`

and `P(B)`

. Tying this all together, we can rewrite `P(A|B)`

as follows

`P(A|B)`

= P(A ∩ B) / `P(B)`

This presents us with another way of computing conditional probabilities: we can compute the probability of the intersection.