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Question #41889

Difficulty: 8/10 | 🟢 Free

The Question

Let E1,E2,E3,…E_1, E_2, E_3, \dots be a countably infinite set of mutually exclusive and exhaustive events. If the probability of occurrence of event EkE_k is given by P(Ek)=λ∣k−4∣3kP(E_k) = \lambda \frac{|k - 4|}{3^k} for all k∈Nk \in \mathbb{N}, then the value of λ\lambda is

Options

  • A.

    108137\frac{108}{137}

  • B.

    45\frac{4}{5}

  • C.

    108133\frac{108}{133}

  • D.

    2734\frac{27}{34}

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

Exhaustive Events Concept|Sigma Notation Properties|Modulus Function Breakdown

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of probability axioms (exhaustive events summing to 1) and the evaluation of infinite arithmetico-geometric series containing a modulus.

The primary trap lies in the modulus function ∣k−4∣|k-4|.

Students frequently ignore the piecewise nature of the modulus, either dropping it entirely or assuming ∣k−4∣=4−k|k-4| = 4-k for all k∈Nk \in \mathbb{N} because the first few terms are positive.

This leads to the incorrect sum ∑(4−k)/3k=5/4\sum (4-k)/3^k = 5/4, yielding the trap answer 4/54/5.

Correct execution requires splitting the domain at the non-differentiable turning point k=4k=4 and evaluating the resulting finite and infinite series separately.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterProbability - Class 11
ConceptExhaustive Events Concept
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree