Prove that $P(A \cup B') = 1 − P(B \cap A')$ I know that $P(A\cup B') = P(A) + P(B') - P(A\cap B')$ and $P(A) = P(A \cap B) + P(A \cap B')$ but still am unable to figure this out. Is there a way to rewrite $P(B ∩ A')$?

$$P(A\cup B') = 1-P\left((A\cup B')'\right) = 1-P\left(A'\cap B\right)$$ as $(A\cup B')' = (A'\cap B)$.
