Since A and B are independent events
∴ P(A ∩ B) = P(A) P(B) ...(1)
P(at least one of A and B) = P(A ∪ B)
= P(A) + P(B) - P(A ∩ B)
= P( A) + P(B) - P(A) P(B) [∵ of(1)]
= P(A) + P(B) [1 - P(A)]
= P(A) + P(B). P(A') = 1 - P(A') + P(B) P(A')
= 1 - P(A') [1 - P(B)] = 1 - P(A') P(B')