If A and B are events such that $P(A' ∪B')=\frac{1}{3} $ and $P(A ∪ B)=\frac{4}{9}$ then the value of $P(A')+P(B')$ is :

$\frac{8}{9}$

The correct answer is Option (3) → $\frac{8}{9}$

P(A′ ∪ B′) = 1/3

Using De Morgan’s Law: A′ ∪ B′ = (A ∩ B)′

P((A ∩ B)′) = 1/3
⇒ 1 − P(A ∩ B) = 1/3
⇒ P(A ∩ B) = 2/3

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

4/9 = P(A) + P(B) − 2/3

4/9 = P(A) + P(B) − 6/9
⇒ P(A) + P(B) = 10/9

Finally,
P(A′) + P(B′) = (1 − P(A)) + (1 − P(B))
= 2 − (P(A) + P(B))
= 2 − 10/9
= 8/9