Question

Let \( A_1, A_2 \) and \( A_3 \) be three events such that \[ P(A_i) = \frac{1}{3}, i = 1, 2, 3; P(A_i \cap A_j) = \frac{1}{6}, 1 \leq i \neq j \leq 3 \text{and} P(A_1 \cap A_2 \cap A_3) = \frac{1}{6}. \] Then the probability that none of the events \( A_1, A_2, A_3 \) occur equals ...........

Hint:

For problems involving multiple events, use the principle of inclusion-exclusion to find the probability of their union.

Answer 1

Correct Answer: 0.3 - 0.4

Step 1: Use the principle of inclusion-exclusion. 
To calculate the probability that none of the events \( A_1, A_2, A_3 \) occur, we first calculate the probability that at least one of them occurs. Using the inclusion-exclusion principle: \[ P(A_1 \cup A_2 \cup A_3) = P(A_1) + P(A_2) + P(A_3) - P(A_1 \cap A_2) - P(A_1 \cap A_3) - P(A_2 \cap A_3) + P(A_1 \cap A_2 \cap A_3). \] Substitute the given values: \[ P(A_1 \cup A_2 \cup A_3) = 3 \times \frac{1}{3} - 3 \times \frac{1}{6} + \frac{1}{6} = 1 - \frac{1}{2} + \frac{1}{6} = \frac{2}{3}. \]

Step 2: Calculate the probability that none of the events occur. 
The probability that none of the events occur is the complement of the probability that at least one occurs: \[ P(\text{none of } A_1, A_2, A_3 \text{ occur}) = 1 - P(A_1 \cup A_2 \cup A_3) = 1 - \frac{2}{3} = \frac{1}{3}. \]

Step 3: Conclusion. 
The correct answer is \( 0.33 \).