Question

If $ P[E_1] = P_1 $ and $ E_1 $ and $ E_2 $ are mutually exclusive, then $ P[\text{neither } E_1 \text{ nor } E_2] $ is equal to

• A. \( (1 - P_1)(1 - P_2) \)
• B. \( 1 - (P_1 + P_2) \)
• C. \( P_1 + P_2 - 1 \)
• D. None of these

Hint:

For mutually exclusive events, the probability of the union of events is the sum of their individual probabilities.

Answer 1

The Correct Option is B

Since \( E_1 \) and \( E_2 \) are mutually exclusive, the probability that neither event occurs is the complement of the probability that at least one of the events occurs.
This can be expressed as: \[ P[\text{neither } E_1 \text{ nor } E_2] = 1 - P[E_1 \cup E_2] \] Since \( E_1 \) and \( E_2 \) are mutually exclusive, we know: \[ P[E_1 \cup E_2] = P[E_1] + P[E_2] = P_1 + P_2 \] Therefore: \[ P[\text{neither } E_1 \text{ nor } E_2] = 1 - (P_1 + P_2) \] Conclusion Thus, the correct answer is \( 1 - (P_1 + P_2) \).