TRUE

If A and B are independent events. It implies-

P(A ∩ B) = P(A)P(B)

∵ P(A′ ∪ B) = P(A’) + P(B) – P(A’ ∩ B)

and P(A′ ∪ B) represents the probability of event ‘only B’ excluding common points.

From Venn diagram we can see:

∴ P (A′ ∩ B) = P(B) – P (A ∩ B)

⇒ P (A′ ∪ B) = P(A’) + P(B) – P(B) + P (A ∩ B)

⇒ P (A′ ∪ B) = 1 – P(A) + P(A)P(B) {independent events}

⇒ P(A′ ∪ B) = 1 – P(A){1 – P(B)}

⇒ P(A′ ∪ B) = 1 – P(A)P(B’) …proved