Let $A, B$ be random variables. Does $P(A|B)=1-P(\overline{A}|B)$? Say I have discrete random variables $A$ and $B$. Is it true that $P(A|B)=1-P(\overline{A}|B)$?
My intuition is conflicting; it makes sense that if $P(A|B)=p$, then it is only possible that $P(\overline{A}|B)=1-p$ and vice versa, since the probability of not A given B and the probability of A given B must add to 1 since there are no other options. 
However, my intuition is usually wrong with probability, things are rarely as they seem. Can anyone give me insight into this?
 A: Yes this is true. If B is the condition that must occur in both cases, then it sort of becomes irrelevant, because cases where B does not occur can be ignored. In this case, we can look at the space of outcomes where B occurs, within which there are two options: A and not A. Therefore these probabilities must add to 1.
A: I like the other answer (+1) as it is simple and based on intuition. This can also be demonstrated using relatively simple probability rules. 
\begin{align*}
P(\bar A | B) &= \frac{P(\bar A\cap B)}{P(B)} \\[1.2ex]
&= \frac{P(B) - P(A\cap B)}{P(B)} \\[1.2ex]
&= 1 - \frac{P(A\cap B)}{P(B)} \\[1.2ex]
&= 1 - P(A|B) \\[1.2ex]
&\square
\end{align*}

Update to address question in the comments.
Claim: If $A$ and $B$ are conditionally independent given $C$, then $P(A|B\cap C) = P(A|\bar{B}\cap C)$. 
Again, this can be demonstrated using only simple probability rules. The idea is, once we know that $C$ occurred, the probability of $A$ "doesn't care" whether or not $B$ occurred. So we will demonstrate this by showing both the LHS and RHS are equal to $P(A|C)$. 
\begin{align*}
P(A|B\cap C) &= \frac{P(A\cap (B\cap C))}{P(B\cap C)} \\[1.2ex]
&= \frac{P(A\cap B | C)P(C)}{P(B\cap C)} \\[1.2ex]
&= \frac{P(A|C)P(B|C)P(C)}{P(B|C)P(C)} && \text{(conditional independence)} \\[1.2ex]
&= P(A|C)
\end{align*}
The RHS is slightly trickier. We use the fact $P(E\cap \bar F) = P(E) - P(E\cap F)$ several times. 
\begin{align*}
P(A|\bar B\cap C) &= \frac{P(A\cap (\bar B\cap C))}{P(\bar B\cap C)} \\[1.2ex]
&= \frac{P(A\cap C) - P(A\cap B \cap C)}{P(C) - P(B\cap C)} \\[1.2ex]
&= \frac{P(A|C)P(C) - P(A\cap B | C)P(C)}{P(C) - P(B|C)P(C)} \\[1.2ex]
&= \frac{P(A|C) - P(A|C)P(B|C)}{1 - P(B|C)} && \text{(conditional independence)} \\[1.2ex]
&= \frac{P(A|C)(1-P(B|C)}{(1-P(B|C)} \\[1.2ex]
&= P(A|C)
\end{align*}
