# Deriving Bayes Rule from conditional probability [duplicate]

Bayes Rule and Conditional Probability look so similar to me. I'm having a hard time figuring out how to derive Bayes from the conditional probability equation. If I start with $$P(A,B) = P(A|B)P(B)$$, how do I get to $$P(A|B) = \frac{P(B|A)P(A)}{P(B)}$$

• Use conditional probability twice to write P(A,B) as equal to two different things. Aug 30, 2021 at 2:57
• (I didn't post an explicit solution due to the impression that this was very likely a self-study question) Aug 30, 2021 at 11:58

$$\mathbb{P}(A,B) = \mathbb{P}(B|A) \mathbb{P}(A) = \mathbb{P}(A|B) \mathbb{P}(B).$$
\begin{align} \mathbb{P}(A|B) = \frac{\mathbb{P}(A,B)}{\mathbb{P}(B)} = \frac{\mathbb{P}(B|A) \mathbb{P}(A)}{\mathbb{P}(B)}. \end{align}