I am to find $P(A \cup B')$ with the following information: $P(A) = .3$, $P(B) = .5$, $P(A \cup B) = .68$
I drew a diagram for $P(A \cup B')$ which appeared to include everything except the part of $B$ which is not shared by $A$. So, I found the probability to be $1-P(B) + P(A \cap B) = .38$
Is this right?
Second, and the problem with which I am struggling more, is finding $P(B)$ and $P(B \mid A)$ with the following information:
$P(A \mid B) = 0.2$, $P(A \mid B') = 0.3$, and $P(A) = 0.22$
I have tried to use several formulas but to no avail. Something is telling me that from the conditional probabilities there is a way to find $P(B)$. I have tried product rule formulas and De Morgan's law, and I googled formulas but I can't narrow it down to 1 unknown to solve. I found the ratio of $P(B)$ to $P(B \mid A)$, but no answers. Any tips would be appreciated.
HINT
According to the inclusion-exclusion principle, it results that \begin{align*} \mathbb{P}(A\cup B^{c}) & = \mathbb{P}(A) + \mathbb{P}(B^{c}) - \mathbb{P}(A\cap B^{c})\\\\ & = \mathbb{P}(A) + 1 - \mathbb{P}(B) + \mathbb{P}(A\cap B^{c}) \end{align*}
Moreover, due the finite additivity property, we do also have \begin{align*} \mathbb{P}(A) = \mathbb{P}(A\cap B^{c}) + \mathbb{P}(A\cap B) \Rightarrow \mathbb{P}(A\cap B^{c}) = \mathbb{P}(A) - \mathbb{P}(A\cap B) \end{align*}
Finally, it remains to determine $\mathbb{P}(A\cap B)$. This can be done as follows: \begin{align*} \mathbb{P}(A\cup B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A\cap B) \Rightarrow \mathbb{P}(A\cap B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A\cup B) \end{align*}
Gathering all the known information, you can obtain the desired result.
Can you take it from here?