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Question #41013

Difficulty: 8/10 | 🟢 Free

The Question

Let $A, B, C$ be three events in a probability space such that $A$ and $C$ are mutually exclusive. The events $A$ and $B$ are independent, and $B$ and $C$ are independent. Given that $P(A \cup B) = \frac{3}{4}$ , $P(B \cup C) = \frac{5}{8}$ , and $P(A \cup C) = \frac{3}{4}$ . Let $p_1$ be the probability that exactly one of the events $A, B, C$ occurs, and $p_2$ be the probability that exactly two of the events $A, B, C$ occur. The value of $\frac{p_1}{p_2}$ is:

Options

A.
$\frac{4}{3}$
B.
$\frac{7}{5}$
C.
$\frac{5}{3}$
D.
$\frac{1}{2}$

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

Mutually Exclusive vs Independent|Sample Space Partitioning|Venn Diagram Overlap Logic

Why This Question Is Tricky

Common Mistakes

This question tests the lethal conflation of pairwise independence with mutual independence, and the misapplication of the product rule to mutually exclusive events.

Students often mechanically assume $P(A \cap C) = P(A)P(C)$ because $A,B$ and $B,C$ are independent, leading to an incorrect $p_2 = 5/16$ and $p_1 = 7/16$ .

Furthermore, the problem forces students to algebraically decode the probabilities first, punishing those who cannot translate set-theoretic unions into algebraic equations.

The core trap occurs when calculating $P(A \cap B \cap C)$ : students either assume it is $P(A)P(B)P(C)$ or struggle to deduce that $A \cap C = \emptyset \implies A \cap B \cap C = \emptyset$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 11
Concept	Mutually Exclusive vs Independent
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free