Is this conditional probability statement true?

Question

Does it come from Bayes theorem that $P(¬A|B) = 1 - P(A|B)$ ? Because $P(¬A|B) = \frac{P(¬AB)}{P(B)}$ and $P(A|B) = \frac{P(AB)}{P(B)}$. So $P(¬AB)$ here is equal to $1-P(AB)$ or no?

Answer 1

$\mathsf P(\neg A\mid B)$ is a probability mass function.   As such, it definitely follows the Law for Probability of Complements.

$\therefore~~\mathsf P(\neg A\mid B)=1-\mathsf P(A\mid B)$

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This can be demonstrated using by Bayes' Theorem$^\dagger$, and the Law of Additivity (for Probabilities of Disjoint Unions).   Since $B$ can be partitioned into such events, therefore: $\mathsf P(B)=\mathsf P(\neg A\cdot B)+\mathsf P(A\cdot B)$.  So:-

$$\begin{align}\mathsf P(\neg A\mid B)&=\dfrac{\mathsf P(\neg A\cdot B)}{\mathsf P(B)}\\[2ex]&=\dfrac{\mathsf P(B)-\mathsf P(A\cdot B)}{\mathsf P(B)}\\[2ex]&=1-\mathsf P(A\mid B)\end{align}$$

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$~\\~$

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$\dagger$ Rather, that is using the Definition of Conditional Probability, to be fully correct.   $\mathsf P(X\mid Y)\mathop{:=}\mathsf P(X\cdot Y)/\mathsf P(Y)$ when event $Y$ has an non-zero probability measure.