What is $P(B|A)$ given the following information? Can someone verify my solution? It doesn't seem correct. Mostly confused by the first part of the problem.
Suppose we have two events $A$ and $B$. The probability for one and only one event of $A$ and $B$ is $0.3$, $P(A) = 0.2$ and $P(B) = 0.3$. What is $P(B|A)$?
It is not known whether  $A$  and $B$ are independent or dependent events.
My work:
Recall $P(B|A) = \dfrac{P(A \cap B)}{P(A)}$. The intersection between two events $A$ and $B$ is the product of their probabilities. $0.2 \times 0.3 = 0.06$. So, $P(B|A) = \dfrac{0.06}{0.2} \implies P(B|A) = 0.3$
 A: Taking $P(A \cap B) = P(A)P(B)$ is assuming that $A$ and $B$ are independent events, which is explicitly not given in the question. Observe that, by your logic,
$$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{P(A)P(B)}{P(A)} = P(B)$$
For any events $A$ and $B$. This is clearly false - you should easily be able to come up with an intuitive example of two events $A$ and $B$ such that $P(B|A) \neq P(B)$
What you instead need to use is that the probability of either $A$ or $B$ occurring but not both occurring is $0.3$. You can notice that the probability of $A$ or $B$ occurring is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, by definition. Then, to get the times where one occurs but not both, we want to exclude the times where both $A$ and $B$ occur, so we subtract $P(A \cap B)$, which is by definition the probability of both $A$ and $B$ occurring. Therefore,
\begin{align*}P(A \textrm{ or } B \textrm{ but not both}) &= P(A \cup B) - P(A \cap B) \\
&= P(A) + P(B) - P(A \cap B) - P(A \cap B) \\
0.3 &= 0.2 + 0.3 - 2P(A \cap B) \\
-0.2 &= -2P(A \cap B) \\
P(A \cap B) &= 0.1 \end{align*}
Then, you just plug this in!
$$P(B|A) = \frac{0.1}{0.2} = \frac{1}{2}$$
And that's your final answer!
