If A and B are two events such that P(A) = (1)/(3), P(B) = (1)/(5)...

Given,

$P\left(A\right) = \dfrac{1}{3}, P\left(B\right) = \dfrac{1}{5}$ and $P\left(A \cup B\right) = \dfrac{1}{2}$

We know that,

$P\left(A \cup B\right) = P\left(A\right) + P\left(B\right) - P\left(A \cap B\right)$

$⇒ P\left(A \cap B\right) = \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{2} = \dfrac{1}{30}$

Now $P\left(A \cap {B}^{'}\right) = \text{Only A} = \dfrac{1}{3} - \dfrac{1}{30} = \dfrac{9}{30}$

And $P\left(B \cap {A}^{'}\right) = \text{Only B} = \dfrac{1}{5} - \dfrac{1}{30} = \dfrac{5}{30}$

Now, $P\left(A ∣ {B}^{'}\right) + P\left(B ∣ {A}^{'}\right) = \dfrac{P\left(A \cap {B}^{'}\right)}{P\left({B}^{'}\right)} + \dfrac{P\left(B \cap {A}^{'}\right)}{P\left({A}^{'}\right)}$

$= \dfrac{\dfrac{9}{30}}{\dfrac{4}{5}} + \dfrac{\dfrac{5}{30}}{\dfrac{2}{3}} = \dfrac{5}{8}$