Correct Answer – Option 3 : 1.1

Concept:

we know that,

\(\rm P(A) +P(B) = P(A \cup B) + P(A \cap B)\)

P(A’) = 1 – P(A) 

Calculations:

Given, For any two events A and B, the probability that at least one of them occur is 0.6

⇒ \(\rm P(A \cup B) = 0.6\)

and A and B occur simultaneously with a probability 0.3

⇒ \(\rm P(A \cap B) = 0.3\)

we know that,

\(\rm P(A) +P(B) = P(A \cup B) + P(A \cap B)\)

⇒ \(\rm P(A) +P(B) = 0.6 + 0.3 = 0.9\)

Also, we know that 

P(A’) + P(B’) = 1 – P(A) + 1 – P(B) 

⇒ P(A’) + P(B’) = 2 – 0.9

⇒ P(A’) + P(B’) = 1.1

Hence, For any two events A and B, the probability that at least one of them occur is 0.6. If A and B occur simultaneously with a probability 0.3, then P(A’) + P(B’) is 1.1