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 $P(B ∩ A')$?

Hint for using your original approach:
Your first equation is relevant:
\begin{equation}
P(A\cup B') = P(A) + P(B') - P(A \cap B')
\end{equation}
Now use your second equation for $B$ as well as $A$
\begin{equation}
P(B) = P(B \cap A) +  P(B \cap A')
\end{equation}
Along with the simple fact that $P(B) + P(B') = 1$, this should be enough for you to prove the desired statement.
