Regarding the case in the above problem where B is a subset of A:
Intuitively I see that the $P(A \cup B)= P(A)$ in this case. However when I compute the$ P(A \cup B)$ I get $1/3$. Where is my logic incorrect?
$P(A \cap B)$ = $P(B|A)\times P(A) = 1 \times 1/2$.
$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 1/3$.
Thanks.
If $B\subseteq A$ then $A\cap B=B$, so $$P(A\cap B)=P(B)=\frac{1}{3}.$$
In your calculation, you have used the wrong value of $P(B|A)$. You should have
$$P(B|A)=\frac{P(B\cap A)}{P(A)}=\frac{P(B)}{P(A)}=\frac{1/3}{1/2}=\frac{2}{3}.$$