Is $P(AB) = P(B \mid A) P(A)$ an error, or am I misunderstanding something? I am currently studying the textbook In all Likelihood by Yudi Pawitan. The section Inverse probability: the Bayesians of chapter 1.4 The emergence of statistics says the following:

The simplest form of the Bayes theorem for two events $A$ and $B$ is
$$P(A \mid B) = \dfrac{P(AB)}{P(B)} = \dfrac{P(B \mid A)P(A)}{P(B \mid A) P(A) + P(B \mid \bar{A})P(\bar{A})} \tag{1.1}$$

I don't think $P(AB) = P(B \mid A) P(A)$ is correct? Is $P(AB) = P(B \mid A) P(A)$ an error, or am I misunderstanding something?

Conditional probability is defined as
$$
P(A|B) = \frac{P(A,B)}{P(B)}
$$
$P(A,B) = P(A|B) P(B)$ is just multiplying both sides by $P(B)$. This is what they use. Also notice that when $A$ and $B$ are binary $P(A,B) = P(AB)$, since $AB=1$ only when both $A$ and $B$ are equal to $1$, so maybe they just made a shortcut here.
