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Question #41050

Difficulty: 8/10 | 🟢 Free

The Question

An experiment has a countably infinite sample space $S = \{E_1, E_2, E_3, \dots\}$ . The probability of each elementary event $E_n$ is given by $P(E_n) = \frac{\alpha}{2^n} + \frac{\beta \cdot 3^n}{(3^n - 1)(3^{n+1} - 1)}$ where $\alpha$ and $\beta$ are positive constants. If $\lim_{n \to \infty} 2^n P(E_n) = \frac{1}{4}$ , then the probability of the event $E_2$ is:

Options

A.
$\frac{5}{26}$
B.
$\frac{529}{2704}$
C.
$\frac{53}{416}$
D.
$\frac{1}{16}$

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

Axiomatic Probability Constraints|Infinite GP Convergence|The Telescopic Series Trap

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to apply the axiomatic probability constraint $\sum P(E_n) = 1$ to a countably infinite sample space, while navigating the 'Telescopic Series Trap'.

The sequence contains a term that superficially resembles a Geometric Progression due to the exponential components.

Students who blindly apply the infinite GP sum formula to the second term will miscalculate its sum as $39/160$ instead of the correct telescopic sum of $1/4$ .

This leads to an incorrect value for $\beta$ and violates the fundamental probability normalization axiom.

Recognizing the hidden telescopic difference $\frac{1}{2}(\frac{1}{3^n-1} - \frac{1}{3^{n+1}-1})$ is crucial for finding the correct distribution parameters.

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