If $$A$$ and $$B$$ are two events such that $$P(A) = \dfrac{1}{3}$$, $$P(B) = \dfrac{1}{5}$$ and $$P(A \cup B) = \dfrac{1}{2}$$, then $$P\left(\dfrac{A}{B'}\right) + P\left(\dfrac{B}{A'}\right)$$ is equal to

We are given $$P(A) = \dfrac{1}{3}$$, $$P(B) = \dfrac{1}{5}$$, and $$P(A \cup B) = \dfrac{1}{2}$$.

First, we find $$P(A \cap B)$$ using the addition rule: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$.

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

$$\dfrac{1}{2} = \dfrac{5 + 3}{15} - P(A \cap B) = \dfrac{8}{15} - P(A \cap B)$$

$$P(A \cap B) = \dfrac{8}{15} - \dfrac{1}{2} = \dfrac{16 - 15}{30} = \dfrac{1}{30}$$

We also need $$P(A') = 1 - \dfrac{1}{3} = \dfrac{2}{3}$$ and $$P(B') = 1 - \dfrac{1}{5} = \dfrac{4}{5}$$.

Now we compute $$P(A \mid B')$$. By definition, $$P(A \mid B') = \dfrac{P(A \cap B')}{P(B')}$$.

$$P(A \cap B') = P(A) - P(A \cap B) = \dfrac{1}{3} - \dfrac{1}{30} = \dfrac{10 - 1}{30} = \dfrac{9}{30} = \dfrac{3}{10}$$

$$P(A \mid B') = \dfrac{3/10}{4/5} = \dfrac{3}{10} \times \dfrac{5}{4} = \dfrac{15}{40} = \dfrac{3}{8}$$

Next, $$P(B \mid A') = \dfrac{P(B \cap A')}{P(A')}$$.

$$P(B \cap A') = P(B) - P(A \cap B) = \dfrac{1}{5} - \dfrac{1}{30} = \dfrac{6 - 1}{30} = \dfrac{5}{30} = \dfrac{1}{6}$$

$$P(B \mid A') = \dfrac{1/6}{2/3} = \dfrac{1}{6} \times \dfrac{3}{2} = \dfrac{3}{12} = \dfrac{1}{4}$$

Therefore, $$P(A \mid B') + P(B \mid A') = \dfrac{3}{8} + \dfrac{1}{4} = \dfrac{3}{8} + \dfrac{2}{8} = \dfrac{5}{8}$$.

The answer is Option B: $$\dfrac{5}{8}$$.