If $P(B) = \frac{3}{5},P(A/B) = \frac{1}{2}$ and $P(A \cup B) = \frac{4}{5}$, then
$P{(A \cup B)^\prime } + P\left( {{A^\prime } \cup B} \right)$ is equal to

Here, $P(B) = \frac{3}{5}$, $P(A/B) = \frac{1}{2}$
and $P(A \cup B) = \frac{4}{5}$

Since, $P(A/B) = \frac{{P(A \cap B)}}{{P(B)}}$
$\Rightarrow$ $P(A \cap B) = P(A/B) \cdot P(B)$
$= \frac{1}{2} \times \frac{3}{5} = \frac{3}{{10}}$

Also, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$\Rightarrow$ $P(A) = \frac{4}{5} - \frac{3}{5} + \frac{3}{{10}} = \frac{1}{2}$

$\therefore$ $P{(A \cup B)^\prime } = 1 - P(A \cup B) = 1 - \frac{4}{5} = \frac{1}{5}$

and $P\left( {{A^\prime } \cup B} \right) = 1 - P(A - B) = 1 - P\left( {A \cap {B^\prime }} \right)$

$= 1 - P(A) \cdot P\left( {{B^\prime }} \right)$

$= 1 - \frac{1}{2} \cdot \frac{2}{5} = \frac{4}{5}$

$\Rightarrow$ $P{(A \cup B)^\prime } + P\left( {{A^\prime } \cup B} \right) = \frac{1}{5} + \frac{4}{5} = \frac{5}{5} = 1$