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.