A fair coin is flipped three times. Let $A$ be the event that a head occurs in the first flip and $B$ be the event that exactly one head occurs.
a) Find $p(A/B)$
b) Are $A$ and $B$ independent?
a) Now we have $$p(A/B)=\frac{p(A \cap B)}{p(B)}$$ but $p(A \cap B)=p(H,\bar{H},\bar{H})=\frac18$ and $p(B)=\frac38$, so we obtain $p(A/B)=\frac13.$
b) We have, $$p(A)=p(H,H,H)+p(H,H,\bar{H})+p(H,\bar{H},\bar{H})+p(H,\bar{H},H)=\frac12$$ so, $A$ and $B$ are not inndependent.
Is my work correct?
On
We have $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$ where, as you stated, $$P(A \cap B) = P(H, \bar H, \bar H) = \frac{1}{2} \cdot (\frac{1}{2})^2 = \frac{1}{8}$$ and $$P(B) = \binom{3}{1}\cdot\frac{1}{2}\cdot (\frac{1}{2})^2 = \frac{3}{8}$$ which gives $$P(A|B) = \frac{1}{3}$$
we know that $$P(A \cap B) = P(A) \cdot P(B) $$ if $A$ and $B$ are independent, this comes from $$ P(A \cap B) = P(A) \cdot P(B) \Leftrightarrow P(A) = \frac{P(A \cap B)}{P(B)} \Leftrightarrow P(A) = P(A|B) $$ Which can only be true if $A$ and $B$ are independent.
we calculate $$P(A) = P(H) = \frac{1}{2}$$ and conclude that $$P(A) \cdot P(B) = \frac{3}{16} \neq \frac{1}{8}$$ Q.E.D
$P(A)$ can be calculated much simpler, as it is simply $P(H)=\frac12$, and you should elaborate some more how you got $P(B)$ to equal $\frac38$.
Also, you did not explain why $A$ and $B$ are dependent, you just threw a couple of numbers down. You must explain why those numbers imply that $A$ and $B$ are not independent.