I was asking myself whether, in probability theory, the probability of B given not-A is just equal to 1 minus the probability of B given A. I guess the correct answer is no but I would like to know why.
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$\begingroup$First, from the equations:
\begin{equation*} P(B | !A) = \frac{P(!A \cap B)}{P(B)} \, , \end{equation*}
while
\begin{equation*} 1 - P(B | A) = 1 - \frac{P(A \cap B)}{P(B)} = \frac{P(B) - P(A \cap B)}{P(B)} \, . \end{equation*}
From the equations, the question comes down to whether \begin{equation*} P(!A \cap B) = 1 - P(A \cap B) \, . \end{equation*}
This cannot hold in a couple of cases. If $A$ and $B$ are mutually exclusive/disjoint, for example, then $B \subseteq !A$ so that LHS = $P(B)$, while RHS = 1. Intuitively, the truth of $A$ ($P(B|A)$) means that $B$ must be false, but knowing that $A$ is false ($P(B|!A)$) does not guarantee that $B$ is true.
In fact, the equation above, which is equivalent to your statement, states $B$ is always true ($P(B) = 1$): $!A \cap B$ is the complement of $A \cap B$, i.e. that exactly one of those is true. So if $!A \cap B$ is false, then $A \cap B$ is true, so that $B$ is true. If, on the other hand, $!A \cap B$ is true, then $B$ must be true. Thus, $P(B) = 1$.
$\endgroup$ $\begingroup$The answer on this is: "no".
If e.g. $A,B$ are independent then also $A^{\complement}$ and $B$ are independent, leading to:$$P(B\mid A)=P(B)=P(B\mid A^{\complement})$$
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