If $P(B) = \frac{3}{5}$, $P(A \mid B) = \frac{1}{2}$ and $P(A \cup B) = \frac{4}{5}$, then $P(A \cup B)' + P(A' \cup B)$

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The correct answer is Option (4) → 1 ##

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

and $P(A \cup B) = \frac{4}{5}$

Since, $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$

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

Also, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$\Rightarrow P(A) = \frac{4}{5} - \frac{3}{5} + \frac{3}{10} = \frac{1}{2}$

$∴P(A \cup B)' = 1 - P(A \cup B) = 1 - \frac{4}{5} = \frac{1}{5}$

and $P(A' \cup B) = 1 - P(A - B) = 1 - P(A \cap B')$

$= 1 - P(A) \cdot P(B')$

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

$\Rightarrow P(A \cup B)' + P(A' \cup B) = \frac{1}{5} + \frac{4}{5} = \frac{5}{5} = 1$