Question

If \( A \) and \( B \) are two events having probabilities, \( P(A) = 0.6 \), \( P(B) = 0.3 \), and \( P(A \cap B) = 0.2 \), then the probability that neither \( A \) nor \( B \) occurs is ______

• A. 0
• B. 0.3
• C. 0.7
• D. 0.8

Hint:

Use the inclusion-exclusion principle to calculate the probability of the union of events and subtract from 1 to find the probability of neither event occurring.

Answer 1

The Correct Option is C

The probability that neither \( A \) nor \( B \) occurs is given by: \[ P(\text{neither } A \text{ nor } B) = 1 - P(A \cup B) \] Using the inclusion-exclusion principle for \( P(A \cup B) \): \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.6 + 0.3 - 0.2 = 0.7 \] Therefore, the probability that neither \( A \) nor \( B \) occurs is: \[ P(\text{neither } A \text{ nor } B) = 1 - 0.7 = 0.3 \]