If $P(A) = \frac{2}{5}$, $P(B) = \frac{3}{{10}}$ and $P(A \cap B) = \frac{1}{5}$, then $P\left( {{A^\prime }/{B^\prime }} \right) \cdot P\left( {{B^\prime }/{A^\prime }} \right)$ is equal to

Here, $P(A) = \frac{2}{5},P(B) = \frac{3}{{10}}$ and $P(A \cap B) = \frac{1}{5}$

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

$= \frac{{1 - [P(A) + P(B) - P(A \cap B)]}}{{1 - P(B)}}$
$= \frac{{1 - \left( {\frac{2}{5} + \frac{3}{{10}} - \frac{1}{5}} \right)}}{{1 - \frac{3}{{10}}}}$

$= \frac{{1 - \left( {\frac{{4 + 3 - 2}}{{10}}} \right)}}{{\frac{7}{{10}}}} = \frac{{1 - \frac{1}{2}}}{{\frac{7}{{10}}}} = \frac{5}{7}$

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

$= \frac{{1 - \frac{1}{2}}}{{1 - \frac{2}{5}}} = \frac{{1/2}}{{3/5}} = \frac{5}{6}$

$\therefore$ $P\left( {{A^\prime }/{B^\prime }} \right) \cdot P\left( {{B^\prime }/{A^\prime }} \right) = \frac{5}{7} \cdot \frac{5}{6} = \frac{{25}}{{42}}$