Is $P(A|B) + P(A|B^C) = 1$? I'm trying to prove / disprove the following statement:

Let A and B be two events. Is it then true that $P(A|B) + P(A|B^c) = 1$? Give
proof or counterexample.

Strategy: I'll try to present a proof, and see what types of assumptions are made along the way in order to arrive for what is sought. Maybe then, after analyzing the assumptions made, I'll be able / not be able, to find a counterexample.
Solution:
Using the fact that $$P(A|B) = \frac{P(A\cap B)}{P(B)}$$ we can rewrite our expression $P(A|B) + P(A|B^c) = 1$ as a single fraction, namely:
$$\frac{P(A\cap B)+P(A\cap B^c)}{P(B)}$$
Now, since $(A\cap B) \cap (A\cap B^c) = \emptyset$, which can very easily be verified by drawing a venn diagram and checking for three possible scenarios, namely when

*

*1: $A \subseteq B$

*2: $A \cap B = \emptyset$

*3:$A \cap B \neq \emptyset$ and $A \nsubseteq B$
Sketching the three scenarios shows us that indeed this is true. Which means we can transform our addition in the numerator into the following:
$$P(A\cap B)+P(A\cap B^c) = P((A\cap B) \cup (A\cap B^c))$$
Applying the distributive law for sets we get:
$$P((A\cap B) \cup (A\cap B^c)) = P((B^c \cup B) \cap A)$$
If we choose $S$ as our sample space, then $B^c \cup B = S$, and since A is an event in our sample space, $S \cap A = A$ itself, meaning that we get:
$$ P(A) / P(B) = 1$$ which is not always true. Hence, disproved.
$\blacksquare$.
Please feel free to give any form of feedback.
Thanks.
 A: Your second equation is wrong. Notice that
$$P(A|B^c)=\frac{P(A\cap B^c)}{P(B^c)},$$
and not
$$P(A|B^c)=\frac{P(A\cap B^c)}{P(B)}.$$
Therefore you can't add the probabilities to get
$$\frac{P(A\cap B)+P(A\cap B^c)}{P(B)}.$$
Anyway, you can take the following example: Let $A$ be the event of getting 2 or 3 from the cube, and $B$ getting 1,2 or 3.
It is easy to calculate that $P(A|B)=\frac23$ and $P(A|B^c)=0$, and thus they don't sum up to 1.
A: One simple counterexample:
Roll a fair die.
A=number is 2
B=number is even
P(A/B)=1/3
P(A/B’)=0
A: Here is an aside comment. A very similar relation related to the proposed one is true:
\begin{align*}
\textbf{P}(A|B) + \textbf{P}(A^{c}|B) = 1
\end{align*}
That is because the conditional probability function is a probability measure as well.
In other terms, one has that
\begin{align*}
\textbf{P}(A|B) + \textbf{P}(A^{c}|B) & = \frac{\textbf{P}(A\cap B)}{\textbf{P}(B)} + \frac{\textbf{P}(A^{c}\cap B)}{\textbf{P}(B)}\\\\
& = \frac{\textbf{P}(A\cap B) + \textbf{P}(A^{c}\cap B)}{\textbf{P}(B)}\\\\
& = \frac{\textbf{P}((A\cap B)\cup(A^{c}\cap B))}{\textbf{P}(B)}\\\\
& = \frac{\textbf{P}((A\cup A^{c})\cap B)}{\textbf{P}(B)}\\\\
& = \frac{\textbf{P}(\Omega\cap B)}{\textbf{P}(B)}\\\\
& = \frac{\textbf{P}(B)}{\textbf{P}(B)} = 1
\end{align*}
Hopefully this contributes!
