Official solution
Given P(A) = \[{1 \over 3}\] and P(B) = \[{1 \over 6}\]
A and B are independent
So P(A\[ \cap \]B) = \[{1 \over 3} \times {1 \over 6}\] = \[{1 \over {18}}\]
P(A\[ \cup \]B) = P(A) + P(B) -- P(A \[ \cap \] B)
= \[{1 \over 3}\] + \[{1 \over 6}\] - \[{1 \over {18}}\]
= \[{4 \over 9}\]
\[P\left( {{A \over {B'}}} \right)\]
= \[{{P\left( {A \cap B'} \right)} \over {P\left( {B'} \right)}}\]
= \[{{P\left( A \right) - P\left( {A \cap B} \right)} \over {1 - P\left( B \right)}}\]
= \[{{{1 \over 3} - {1 \over {18}}} \over {1 - {1 \over 6}}}\] = \[{{1 \over 3}}\]