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 $P(B ∩ A')$?

$P(A \cup C) = 1 - P(A^* \cap C^*)$.
Set $C = B^*$.
