First, let’s calculate $ \Pr(A \cap B) $
Since $A$ and $B$ are independent, $ \Pr(A \cap B) = \Pr(A) \cdot \Pr(B) $
$ \Pr(A \cap B) = \frac{1}{3} \cdot \frac{1}{2} $
$ \Pr(A \cap B) = \frac{1}{6} $
Now, let’s calculate $ \Pr(A \cup B) $
$ \Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B) $
$ \Pr(A \cup B) = \frac{1}{3} + \frac{1}{2} - \frac{1}{6} $
$ \Pr(A \cup B) = \frac{2}{6} + \frac{3}{6} - \frac{1}{6} $
$ \Pr(A \cup B) = \frac{4}{6} $
$ \Pr(A \cup B) = \frac{2}{3} $