Is this always true: $P(A|B) = 1-P(A^c|B)$?

Question

Does this identity hold for all events?

$$ P(A|B) = 1-P(A'|B) $$

Logically speaking, if the probability of $A$ given $B$ occurred is $X$, shouldn't the probability that $A$ does not occur, $A'$, given $B$, be similarly $1-X$?

There is a related question here.

This is the closest that I could get to proving (or disproving) it:

1. $P(A\cap B)=P(A)-P(A' \cap B)$
2. $P(A|B)P(B)=P(A)-P(A'|B)P(B)$
3. $\therefore P(A'|B)= \frac {P(A)} {P(B)}-P(A|B)$

Are there are certain formulae which can be used to prove this? Or does the identity only hold under certain situations, and if so, what kind of situations?

Thanks.

Answer 1

Does this identity hold for all events?

$P(A\mid B)=1−P(A^\prime\mid B)$

Logically speaking, if the probability of A given B occurred is X, shouldn't the probability that A does not occur, A′, given B, be similarly 1−X?

Yes. The complement rule holds for conditional probabilities.

$$\begin{align}\Pr(B) & = \Pr((A\cap B) \cup (A^\prime\cap B)) & \text{by total probability law} \\ & = \Pr(A\cap B)+\Pr(A^\prime\cap B) & \text{because of mutual exclusion} \\ \implies \Pr(A\cap B) & = \Pr(B)-\Pr(A^\prime\cap B) & \text{by rearangement}\\\therefore \Pr(A\mid B)&=1 - \Pr(A^\prime\mid B) & \text{by division by }\Pr(B)\end{align}$$

Answer 2

The above answer is correct, just for clarification your step 1. in the proof in the question is incorrect.

1. $P(A\cap B) = P(A) - P(A\cap B')$

or

1. $P(A\cap B) = P(B) - P(A'\cap B)$

would be true.