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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 $$ P(A|B) = \frac{P(B|A)P(A)}{P(B)} $$

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

1 Answer 1

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Decomposing the joint probability both ways, you get:

$$\mathbb{P}(A,B) = \mathbb{P}(B|A) \mathbb{P}(A) = \mathbb{P}(A|B) \mathbb{P}(B).$$

Consequently, you get:

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

For more information on the intuition of this result, and a way of expressing it via a symmetry principle, see this related answer.

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