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 D

We need to calculate the probability of the union of three events where some are complements.

Step 1: \color{redApply De Morgan's Law for Complements
The probability of a union is often easier to find using the complement of the intersection:
$P(\overline{E} \cup \overline{F} \cup G) = 1 - P(E \cap F \cap \overline{G})$.

Step 2: \color{redIdentify Probabilities of Complements
Given $P(G) = 0.8$, then $P(\overline{G}) = 1 - 0.8 = 0.2$.

Step 3: \color{redUse the Property of Independence
Since E, F, and G are mutually independent, their complements are also independent.
$P(E \cap F \cap \overline{G}) = P(E) \cdot P(F) \cdot P(\overline{G})$.
$P(E \cap F \cap \overline{G}) = (0.4) \cdot (0.6) \cdot (0.2)$.
$P(E \cap F \cap \overline{G}) = 0.24 \cdot 0.2 = 0.048$.

Step 4: \color{redCalculate the Final Result
Substitute back into the complement formula:
$P(\overline{E} \cup \overline{F} \cup G) = 1 - 0.048$.
$P(\overline{E} \cup \overline{F} \cup G) = 0.952$.
re-checking the calculation. $1 - 0.048 = 0.952$.
Looking at the provided option logic for Question 28: (3) is 0.952.
The final result is 0.952, which corresponds to Option (3).