The probability that at least one of the two events A and B occurs is 0.7, and the probability that both A and B occur is 0.25. Then \(P(A')+P(B')\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1.05
Here \(P(A\cup B) = 0.7\) and \(P(A\cap B) = 0.25\). By the addition theorem:
\(P(A)+P(B) = P(A\cup B)+P(A\cap B)\)
\(P(A)+P(B) = 0.7+0.25 = 0.95\)
Since \(P(A') = 1-P(A)\) and \(P(B') = 1-P(B)\):
\(P(A')+P(B') = 2-\left(P(A)+P(B)\right)\)
\(P(A')+P(B') = 2-0.95 = 1.05\)
Why the other options are incorrect
Option (1) 0.95 — Incorrect. This is P(A) + P(B); the complements have not been taken.
Option (2) 0.30 — Incorrect. This is \(1-0.7 = P(A'\cap B')\), the probability that neither event occurs.
Option (4) 1.55 — Incorrect. This uses P(A) + P(B) = 0.7 − 0.25 = 0.45; the common part must be added to the union, not subtracted.