# In finding P(A|B), we use P(B|A) and vice versa. How do we find either without knowing one already?

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## 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?

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.

FAQ: Bayes' Theorem - Bayes' Theorem