Is it true that $P(A \mid B) + P(A^c\mid B) = 1$?

Question

That is, generally is it true that $P(A \mid B) + P(A^\complement\mid B) = 1$? I'm doing a problem involving Bayes' rule and my professor (seemingly) makes use of this fact.

Any clarity on this would be great!

Answer 1

By definition:

$$P(A|B)+P(A'|B)=\frac{P(A \cap B)}{P(B)}+\frac{P(A' \cap B)}{P(B)}=\frac{P(A \cap B)+P(A' \cap B)}{P(B)}$$

Next $A \cap B$ and $A' \cap B$ are disjoint, so $$P(A \cap B)+P(A' \cap B)=P((A \cap B) \cup A' \cap B)=P((A \cup A')\cap B)=P(B)$$

Finally:

$$\frac{P(A \cap B)+P(A' \cap B)}{P(B)}=\frac{P(B)}{P(B)}=1$$

Answer 2

As far as I know it is. (I'm not an expert in probability though)

The way I understand it is that $P(A \mid B) = \frac{P(A \land B)}{P(B)}$, so then we'd have $$P(A\mid B)+P(\lnot A \mid B) = \frac{P(A \land B)+P((B \land \lnot A)\land B)}{P(B)} = \frac{P((A\land B)\lor((B\land\lnot A)\land B))}{P(B)} = \frac{P(B) - P((B\land\lnot A)\land A))}{P(B)} = \frac{P(B)}{P(B)} = 1.$$

Does this make sense?

Answer 3

Yes, if the two conditional probabilities are defined, they sum to $1$.

This derives from the facts that $A\cap B$ and $A^C\cap B$ are disjoint (where $A^C$ is the complement of the set $A$) and that $(A\cap B)\cup(A^C\cap B)=B.$ So $$(P(A\mid B)+P(A^C\mid B))P(B)=P(A\cap B)+P(A^C\cap B)=P(B).$$