# 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.

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: $$P(A\cup X)= P(A) + P(X) - P(A \cap X)=...$$

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

$$P(A\cup X)= P(A) + P(X) - P(A \cap X) = 0.48 + 0.2 - 0.08 = 0.6$$

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 $$(2)$$ $$P(A\cup X)=0.48+0.12=0.6$$