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Question #41020

Difficulty: 8/10 | 🟢 Free

The Question

A discrete random variable $X$ takes values in the set $S = \{-3, -2, -1, 0, 1, 2, 3\}$ . The probability mass function of $X$ is given by $P(X=k) = C \left( 5 - |k| + \lambda \cdot \text{sgn}(k^3 - k) \right)$ where $C$ and $\lambda$ are real constants, and $\text{sgn}(\cdot)$ denotes the signum function. If the expected value of $X$ is $E(X) = \frac{20}{23}$ , then the probability $P(X \ge 1)$ is equal to

Options

A.
$\frac{13}{23}$
B.
$\frac{14}{23}$
C.
$\frac{31}{46}$
D.
$\frac{18}{23}$

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

Equiprobability Fallacy|Probability Mass Function (PMF)|Signum Function Discontinuity

Why This Question Is Tricky

Common Mistakes

This question tests the properties of Probability Mass Functions (PMF) combined with the discontinuity of the signum function.

The core trap lies in evaluating $\text{sgn}(k^3-k)$ .

Students frequently commit the equiprobability/continuity fallacy by assuming $\text{sgn}(x)$ only evaluates to $\pm 1$ , missing that $\text{sgn}(0) = 0$ at the roots $k=-1, 0, 1$ .

If a student incorrectly assumes $\text{sgn}(k^3-k) = \text{sgn}(k)$ , they will miscalculate the expected value sum as $12\lambda/23$ instead of $10\lambda/23$ , leading to an incorrect $\lambda = 5/3$ and the trap answer $14/23$ .

The problem also requires recognizing odd/even function properties over symmetric discrete domains to efficiently find $C$ and $E(X)$ .

Quick Facts About This Question

Question Information

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