A and B are two independent events such that \(P(A) = \displaystyle\frac{1}{3}\) and \(P(B) = \displaystyle\frac{1}{4}\). The probability that neither A nor B occurs is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{2}\)
“Neither A nor B” means A does not occur and B does not occur, that is \(A' \cap B'\). When A and B are independent, A’ and B’ are also independent.
\( \displaystyle P(A') = 1 - \frac{1}{3} = \frac{2}{3} \)
\( \displaystyle P(B') = 1 - \frac{1}{4} = \frac{3}{4} \)
Multiplying:
\( \displaystyle P(A' \cap B') = \frac{2}{3} \times \frac{3}{4} \)
\( \displaystyle = \frac{1}{2} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{12}\) — Incorrect. This is \(\frac{1}{3} \times \frac{1}{4}\), the probability that both A and B occur.
Option (3) \(\displaystyle\frac{5}{12}\) — Incorrect. This is \(1 - \frac{1}{3} - \frac{1}{4}\), which forgets to add back \(P(A \cap B) = \frac{1}{12}\).
Option (4) \(\displaystyle\frac{7}{12}\) — Incorrect. This is \(\frac{1}{3} + \frac{1}{4}\), the sum of \(P(A)\) and \(P(B)\); it neither subtracts \(P(A \cap B)\) nor takes the complement.