Probability Proof $P(B-A) =P(B)-P(A\cap B)$

Question

How would I go about proving this statement:

$P(B-A) =P(B)-P(A\cap B)$

Answer 1

Observe that $(B-A) \cup (A \cap B) = B$ and $(B-A) \cap (A \cap B) = \emptyset.$

So $P(B) = P((B-A) \cup (A \cap B)) = P(B-A) + P(A \cap B).$ Therefore $P(B-A) = P(B) - P(A \cap B).$

Answer 2

Use $P(X)+P(Y)=P(X\cap Y)+P(X\cup Y)$ with $X:=B-A,\,Y:=A\cap B$ so $X\cap Y=\emptyset,\,X\cup Y=B$.