The probability that at least one of the events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.3, then \(P(A') + P(B')\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1.1
“At least one of A and B” is the event \(A \cup B\), and “simultaneously” is \(A \cap B\). So \(P(A \cup B) = 0.6\) and \(P(A \cap B) = 0.3\).
By the addition rule:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( 0.6 = P(A) + P(B) - 0.3 \)
\( P(A) + P(B) = 0.9 \)
Since \(P(A') = 1 - P(A)\) and \(P(B') = 1 - P(B)\):
\( P(A') + P(B') = 2 - [P(A) + P(B)] \)
\( = 2 - 0.9 \)
\( = 1.1 \)