Question

Let E, F and G be mutually independent events such that $P(E) = 0.4$, $P(F) = 0.6$ and $P(G) = 0.8$ then $P(\overline{E} \cup \overline{F} \cup G)$ is

• A. 0.192
• B. 0.048
• C. 0.952
• D. 0.808

Hint:

Always use $1 - P(\text{Complement})$ when dealing with unions of independent events. It converts a complex addition problem into a simple multiplication problem.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:

E, F, G are mutually independent with \( P(E) = 0.4 \), \( P(F) = 0.6 \), \( P(G) = 0.8 \). Find \( P(\overline{E} \cup \overline{F} \cup G) \).

Step 2: Detailed Explanation:

• Take the complement: \( P(\overline{E} \cup \overline{F} \cup G) = 1 - P(E \cap F \cap \overline{G}) \).

• By independence, \( P(E \cap F \cap \overline{G}) = 0.4 \times 0.6 \times 0.2 = 0.048 \).

• Hence the required probability is \( 1 - 0.048 = 0.952 \).

Step 3: Final Answer:
The correct option is (C) 0.952.