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Question #41840

Difficulty: 8/10 | 🟢 Free

The Question

Let $A$ , $B$ , and $C$ be three events in a probability space such that $A$ and $B$ are independent, $B$ and $C$ are independent, and $A$ and $C$ are mutually exclusive. If $P(A \cup B) = \frac{2}{3}$ , $P(B \cup C) = \frac{1}{2}$ , and $P(A \cup C) = \frac{3}{4}$ , then the value of $P(A^c \cap B^c \cap C^c)$ is

Options

A.
$\frac{1}{6}$
B.
$\frac{1}{4}$
C.
$\frac{1}{8}$
D.
$\frac{5}{24}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

De Morgan's Laws in Probability|Independent vs Disjoint Events|Mutually Exclusive vs Independent

Why This Question Is Tricky

Common Mistakes

This question tests the critical distinction between mutually exclusive and independent events, exploiting the 'Product Rule Assumption'.

Students often instinctively calculate $P(A^c \cap B^c \cap C^c)$ as $P(A^c)P(B^c)P(C^c)$ after finding the individual probabilities, yielding the trap answer $1/4$ .

This assumes mutual independence, which is impossible here because $A$ and $C$ are mutually exclusive (and thus highly dependent, meaning their complements are also dependent).

The correct approach requires using De Morgan's Law to invert the intersection into a union, and then applying the Addition Theorem where the non-exclusivity and zero-probability intersections ( $A \cap C = \emptyset \implies A \cap B \cap C = \emptyset$ ) dictate the final result.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 12
Concept	De Morgan's Laws in Probability
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free