If A and B are such that $P\left( {{A^\prime } \cup {B^\prime }} \right) = \frac{2}{3}$ and $P(A \cup B) = \frac{5}{9}$, then $P\left( {{A^\prime }} \right) + P\left( {{B^\prime }} \right)$ is equal to……….

Here, $P\left( {{A^\prime } \cup {B^\prime }} \right) = \frac{2}{3}$ and $P(A \cup B) = \frac{5}{9}$
$P\left( {{A^\prime } \cup {B^\prime }} \right) = 1 - P(A \cap B)$

$\Rightarrow$ $\frac{2}{3} = 1 - P(A \cap B)$
$\Rightarrow$ $P(A \cap B) = 1 - \frac{2}{3} = \frac{1}{3}$

$= 2 - [P(A) + P(B)]$
$= 2 - [P(A \cup B) + P(A \cap B)]$
$= 2 - \left( {\frac{5}{9} + \frac{1}{3}} \right) = 2 - \left( {\frac{{5 + 3}}{9}} \right)$

$= \frac{{18 - 8}}{9} = \frac{{10}}{9}$