Finding $P(\overline{A} \cup B)$ when $P(A)$, $P(B)$ and $P(A \cup B$) is known

Question

Two events $A$ and $B$ are such that $P(A) = 0.2$, $P(B) = 0.3$ and
$P(A \cup B) = 0.4$.
(a) Find $P(A \cap B)$.
(b) Find $P(\overline{A} \cup B)$.

Answer:(a)
Recall that:
$$ P(A) + P(B) – P(A \cap B) = P(A \cup B) $$
We have:
\begin{align*}
-P(A \cap B) &= P(A \cup B) – P(A) – P(B) \\
P(A \cap B) &= -P(A \cup B) + P(A) + P(B) \\
P(A \cap B) &= -0.4 + 0.2 + 0.3 \\
P(A \cap B) &= 0.1 \\
\end{align*}
(b)
\begin{align*}
P(A) + P(\overline{A}) &= 1 \\
P(\overline{A}) &= 1 – P(A) = 1 – 0.1 \\
P(\overline{A}) &= 0.9 \\
P(\overline{A} \cup B) &= P(\overline{A}) + P(B) – P(\overline{A} \cap B)
\end{align*}
Now I am stuck because I do not know how to find $P(\overline{A} \cap B)$.

Now, based upon the comments from lulu, I have:
\begin{align*}
P(A \cap B) + P(\overline{A} \cap B) &= P(B) \\
.1 + P(\overline{A} \cap B) &= .4 \\
P(\overline{A} \cap B) &= 0.3 \\
P(\overline{A} ) &= 0.8 \\
P(\overline{A} \cup B) &= 0.8 + 0.4 – 0.3 \\
P(\overline{A} \cup B) &= 0.9 \\
\end{align*}

Answer 1

Your answers are both correct. Alternatively, a Venn diagram helps with the set algebra.

1. $P(A\cup B)=P(A)+P(B)-P(A\cap B).$
   So, $P(A\cap B)=0.2+0.3-0.4=0.1.$
2. $P(\overline{A} \cup B)=P(B)+P(\overline{A\cup B})=0.3+(1-0.4)=0.9.$