Solution
= Solution
There are eighteen equally likely first choices. Counting the relevant subsets gives
$$
\boxed{P(A)=\frac{13}{18},\qquad P(B)=\frac{10}{18}=\frac59,\qquad
P(A\cap B)=\frac8{18}=\frac49}.
$$
For <conditional probabilities>, restrict the denominator to the conditioning subset:
$$
\boxed{P(A\mid B)=\frac8{10}=\frac45,\qquad P(B\mid A)=\frac8{13}}.
$$
The five counts come from the same finite <sample space>, so no independence assumption is involved.