Does $P(A\cap B) + P(A\cap B^c) = P(A)$?

Question

Based purely on intuition, it would seem that the following statement is true, when thinking of the events as sets:

$$P(A\cap B) + P(A\cap B^c) = P(A)$$

However, I am not sure if this is true, and cannot find out how to prove it, or describe a straight forward intuition of it.

Is this equation true? And if not, is there another way to link $P(A\cap B)$ and $P(A\cap B^c)$ with $P(A)$?

Answer 1

As $A\cap B$ and $A\cap B^c$ are disjoint, we have $$P(A\cap B)+P(A\cap B^c)=P((A\cap B)\cup(A\cap B^c))=B(A\cap(B\cup B^c))=P(A). $$

Answer 2

Yes it is true. In fact is one of the basic properties.

$$ P(A)=P(A\cap\Omega)=P(A\cap\left(B\cup B^c\right))=P((A\cap B)\cup (A \cap B^c))=P(A\cap B)+P(A\cap B^c) $$