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Since, \( P(A \cap B) = P(A) \cdot P(B) \)
It means \( A \) and \( B \) are independent events, so \( A' \) and \( B' \) are also independent. \[ \therefore P(A \cup B)' = P(A' \cap B') = P(A)' \cdot P(B)' \] Alternative Solution: \[ P(A \cup B)' = 1 - P(A \cup B) = 1 - \{ P(A) + P(B) - P(A) \cdot P(B) \} \] \[ = \{ 1 - P(A) \} \cdot \{ 1 - P(B) \} = P(A)' \cdot P(B)' \]