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
