Question

Let \(E_1, E_2\) and \(E_3\) be three events such that \(P(E_1) = \frac{4}{5}, P(E_2) = \frac{1}{2}\) and \(P(E_3) = \frac{9}{10}\). Then, which of the following statements is FALSE?

• A. \(P(E_1 \cup E_2 \cup E_3) \ge \frac{9}{10}\)
• B. \(P(E_2 \cup E_3) \ge \frac{9}{10}\)
• C. \(P(E_1 \cap E_2 \cap E_3) \le \frac{1}{6}\)
• D. \(P(E_1 \cup E_2) \le \frac{4}{5}\)

Hint:

For any events \(A, B\), \(P(A \cup B) \ge \max(P(A), P(B))\). A union cannot have a smaller probability than its individual events.

Answer 1

The Correct Option is C

Step 1: Recall the formula for union of two events.
\[ P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2). \] Since \(P(E_1 \cap E_2) \ge 0\), \[ P(E_1 \cup E_2) \le P(E_1) + P(E_2) = \frac{4}{5} + \frac{1}{2} = \frac{13}{10}. \] However, probability cannot exceed 1. Hence, \(P(E_1 \cup E_2) \le 1\). The lower bound is \( \max(P(E_1), P(E_2)) = \frac{4}{5}\). Thus, \(P(E_1 \cup E_2) \ge \frac{4}{5}\), not \(\le \frac{4}{5}\).
Step 2: Verify others qualitatively.
(A) True, since the union of three events is at least as large as the largest individual probability (\( \frac{9}{10} \)). (B) True, similar reasoning as (A). (C) True, since by Boole’s inequality, intersection probability cannot exceed the smallest individual probability (\( \frac{1}{2} \)).
Step 3: Conclusion.
Option (D) is the only false statement. Final Answer: \[ \boxed{P(E_1 \cup E_2) \le \frac{4}{5} \text{ is FALSE.}} \]