How to find $\Pr(A)$ if: $\Pr(A|B) = 0.2, \Pr(B) = 0.8$ and $\Pr(A|B^c) = 0.3$?

Question

I need help with a simple homework exercise.

How to find $\Pr(A)$ if: $$\Pr(A|B) = 0.2 $$ $$\Pr(B) = 0.8 $$ $$\Pr(A|B^c) = 0.3$$

I found $\Pr(A\cap B)$ and $\Pr(A\cap B^c)$ but I don't know what to do with. I'm stuck.

Answer 1

We have $$\Pr(A)=\Pr(A|B)\Pr(B)+\Pr(A|B^c)\Pr(B^c).\tag{$1$}$$

This is because $A=(A\cap B)\cup (A\cap B^c)$ and the union is disjoint. So $$\Pr(A)=\Pr(A\cap B) +\Pr(A\cap B^c).\tag{$2$}$$ To get from $(2)$ to $(1)$, use the fact that $\Pr(A\cap B)=\Pr(A|B)\Pr(B)$ and $\Pr(A\cap B^c)=\Pr(A|B^c)\Pr(B^c)$

If you found the two probabilities on the right of $(2)$, then $\Pr(A)$ was one step away.