Question

If \( A, B \) and \( C \) are three events of a random experiment with \( P(A) = 0.4 \), \( P(B) = 0.3 \), and \( P(A \cap B) = 0.2 \), then the probability that neither A nor B occurs is:

• A. \textbf{0.5}
• B. 0.15
• C. 0.13
• D. 0.12

Hint:

To find the probability of “neither A nor B,” use the complement rule: \( P(\text{neither}) = 1 - P(A \cup B) \), and apply inclusion-exclusion for \( P(A \cup B) \).

Answer 1

The Correct Option is A

We are given: \[ P(A) = 0.4,\quad P(B) = 0.3,\quad P(A \cap B) = 0.2 \] We want to find the probability that **neither A nor B** occurs, i.e., \[ P(\text{not } A \cup B) = 1 - P(A \cup B) \] From the formula: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Substitute the values: \[ P(A \cup B) = 0.4 + 0.3 - 0.2 = 0.5 \] Hence, \[ P(\text{neither A nor B}) = 1 - 0.5 = \boxed{0.5} \]