For each pair of events A and B, find P(A|B) and P(B|A). I've got a simple problem here, but I just want to ensure that I'm not losing a simple concept.
Relevant equations:
$$P(A|B) = \frac{P(A \cap B)}{P(B)},$$
$$P(A \cap B) = P(A) + P(B) - P(A \cup B)$$
Let $(S,P)$ be a sample space with $S = \{1, 2, 3, 4, 5\}$, and,
$P(1) = P(2) = 0.1$$P(3) = P(4) = 0.2$
$P(5) = 0.4$

For each pair of events $A$ and $B$, find $P(A|B)$ and $P(B|A)$.
a. $A = \{1, 2, 3\}$; $B = \{2, 3, 4\}$
b. $A = \{1, 2, 3\}$; $B = \{4, 5\}$
c. $A = \emptyset$; $B = \{2, 3, 4\}$
d. $A = \{1, 2, 3, 4, 5\}; B = \{4, 5\}$

If I am asked to find the probability of one of these sets, do I just add the values together? Is there any instance I would multiply them; such as when I'm finding $P(A \cap B)$ vs. just $P(A)$ or $P(B)$? Do I need to use Inclusion-Exclusion?
Here's what I found.
a. $P(A|B) = \frac{0.3}{0.5} = \frac{3}{5}$, $P(B|A) = \frac{0.3}{0.4} = \frac{3}{4}$
b. $P(A|B) = 0$, $P(B|A) = 0$
c. $P(A|B) = 0$, $P(B|A) = undefined$
d. $P(A|B) = \frac{0.3}{0.3} = 1$, $P(B|A) = \frac{0.3}{1} = \frac{3}{5}$

Your approach is correct, as are your answers.  However, your solutions in part (d) are incorrect.
Since $A = \{1, 2, 3, 4, 5\}$ is the entire sample space, $P(A) = 1$.  Since $P(4) = 0.2$ and $P(5) = 0.4$ and $B = \{4, 5\}$, $P(B) = P(4) + P(5) = 0.2 + 0.4 = 0.6$.  Since $B \subseteq A$, $P(A \cap B) = P(B)$.  Hence, 
\begin{align*}
P(A \mid B) & = \frac{P(A \cap B)}{P(B)} = \frac{P(B)}{P(B)} = \frac{0.6}{0.6} = 1\\  
P(B \mid A) & = \frac{P(A \cap B)}{P(A)} = \frac{P(B)}{P(A)} = \frac{0.6}{1} = 0.6
\end{align*}
