If $P(X)=0.2$, $P(A|X)=0.4$ and $P(A|X')=0.5$ find $P(A\cup X)$. 
I've been given the information that $P(X)=0.2$, $P(A|X)=0.4$ and $P(A|X')=0.5$.
I need to find $P(A\cup X)$.

So my initial though was use the fact that $P(A\cup X)= P(A) + P(X) - P(A \cap X)$ but i don't know $P(A)$ and $P(X)$ and I'm not sure how to find them or If i need to find then to solve this problem. Any help would be great.
 A: From $$P(A|X) = {P(A\cap X)\over P(X)}\implies P(A\cap X) = 0,08$$
and since $P(X') = 0,8$:$$P(A|X') = {P(A\cap X')\over P(X')}\implies P(A\cap X') = 0,4$$
Notice that $A = (A\cap X)\cup (A\cap X')$ and that $A\cap X$ and $A\cap X'$ are not compatibile, so $$P(A)= P(A\cap X)+P(A\cap X')= 0,48$$ and now use PIE:
$$P(A\cup X)= P(A) + P(X) - P(A \cap X)=...$$
A: You're right to use the identity, but before that, you'll need to calculate the terms on the right hand side first.
$$P(A \cap X) = P(A|X)P(X) = 0.4 \cdot 0.2 = 0.08$$
Use the law of total probability.
$$\begin{aligned}
P(A) &= P(A|X)P(X) + P(A|X')P(X') \\
&= 0.4 \cdot 0.2 + 0.5 \cdot (1-0.2) \\
&= 0.48
\end{aligned}$$
Now, you can use the formula (special case of inclusion-exclusion principle when $n=2$) in the question body to conclude that
$$P(A\cup X)= P(A) + P(X) - P(A \cap X) = 0.48 + 0.2 - 0.08 = 0.6$$
A: Yes $$P(A\cup X)= P(A) + P(X) - P(A \cap X) \tag{1}$$
is a good start. From $P(A|X)=\frac{P(A\cap X)}{P(X)} \Rightarrow P(A\cap X)=P(A|X)\cdot P(X)=0.4\cdot 0.2=0.08$ and 
$$P(A\cup X)= P(A) + P(X) - P(A \cap X)=P(A) + 0.2 - 0.08=P(A) + 0.12 \tag{2}$$
Then you have the law of total probability
$$P(A)=P(A|X)\cdot P(X)+P(A|X')\cdot P(X')=0.08+0.5\cdot(1-0.2)=0.48$$
and from $(2)$
$$P(A\cup X)=0.48+0.12=0.6$$
