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Question #41046

Difficulty: 8/10 | 🟢 Free

The Question

Let $\Omega = \{1, 2, 3, 4, 5, 6, 7\}$ be a sample space equipped with a valid probability measure $P$ . Consider the events $E_1 = \{1, 2, 3, 4\}$ , $E_2 = \{3, 4, 5, 6\}$ , and $E_3 = \{1, 2, 5, 6\}$ . It is given that $E_1, E_2$ , and $E_3$ are pairwise independent events and $P(E_1) = P(E_2) = P(E_3) = p > 0$ . What is the value of $P(\{7\})$ ?

Options

A.
$\frac{1}{7}$
B.
$\frac{1}{4}$
C.
$\frac{1}{8}$
D.
$\frac{1}{2}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Axiomatic Probability Constraints|Independent vs Disjoint Events|Equiprobability Fallacy

Why This Question Is Tricky

Common Mistakes

This question targets the Equiprobability Fallacy and tests strict adherence to Axiomatic Probability Constraints.

Students often assume that a sample space of 7 elements implies a uniform probability distribution ( $P(\{i\}) = 1/7$ ), which acts as a lethal trap here.

By enforcing pairwise independence on non-disjoint, symmetrically overlapping events, the problem dictates a non-uniform measure.

The student must rely on the additivity axiom over disjoint subsets and the definition of independence to algebraically deduce the true probability measure, revealing that the visual symmetry of the sets masks a strictly constrained, non-equiprobable space.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 11
Concept	Axiomatic Probability Constraints
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free