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I am familiar with the definition $P(A \vert B) = \begin{cases} \dfrac{P(B \vert A) P(A)}{P(B)}, & P(B) \neq 0 \\ \text{undefined}, & P(B) = 0 \end{cases}$

However, I just saw another definition for $P(A \vert B)$:

$P(A \vert B) = \begin{cases} \dfrac{P(AB)}{P(B)}, & P(B) \neq 0 \\ \text{undefined}, & P(B) = 0 \end{cases}$

I have never seen this definition before. How is this definition equivalent to the first? Thank you.

Dom Fomello
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2 Answers2

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Your definition does not make sense. You are defining $P(A|B) $ in terms of $P(B|A)$. That cannot be a definition. However the equation is valid and it is a theorem.

The correct definition is $P(A|B)=\frac {P(A\cap B)} {P(B)}$ if $P(B) >0$. Sometimes $P(A \cap B)$ is written as $P(AB)$.

Kavi Rama Murthy
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  • The first definition just is a simple version of Bayes’ theorem, so it’s not a definition per se but it is a statement that holds true. – SuperSjoerdie Feb 17 '20 at 00:36
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$AB$ is shorthand for the intersect of the events $A$ and $B$, that is: $A\cap B$.

The rest is just the definition of conditional probability:

$$\begin{align}\mathsf P(AB)&=\mathsf P(A\cap B)\\&=\mathsf P(A\mid B)~\mathsf P(B)&&\text{when }\mathsf P(B)\neq 0\text{ and so }\mathsf P(A\mid B)\text{ is defined}\\&=\mathsf P(B\mid A)~\mathsf P(A)&&\text{when }\mathsf P(A)\neq 0\text{ and so }\mathsf P(B\mid A)\text{ is defined}\end{align}$$

Graham Kemp
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