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Question #41687

Difficulty: 8/10 | 🟢 Free

The Question

A discrete random variable $X$ takes values in $\{1, 2\}$ with the probability mass function: $P(X=1) = \frac{1}{4} + \frac{1}{2\pi} \arcsin(\sin \alpha)$ $P(X=2) = \frac{3}{4} - \frac{1}{2\pi} \arccos(\cos \alpha)$ where $\alpha \in [0, 2\pi]$ . Let $S$ be the set of all values of $\alpha$ for which this defines a valid probability mass function. For a fixed $\alpha \in S$ , consider the linear programming problem: Maximize $Z = P(X=1) x + P(X=2) y$ subject to: $2x + y \le 6$ $x + 3y \le 9$ $x \ge 0, y \ge 0$ Let $M(\alpha)$ be the maximum value of $Z$ for a given $\alpha$ . Find the minimum value of $M(\alpha)$ over all $\alpha \in S$ .

Options

A.
$\frac{21}{10}$
B.
2
C.
$\frac{9}{5}$
D.
$\frac{9}{4}$

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

Non-Negativity Restriction|Probability Mass Function (PMF)|Principal Value Branch Boundaries

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of probability constraints, inverse trigonometric principal value branches, and linear programming.

The core trap lies in the condition $\sum P(X=x) = 1$ , which simplifies to $\arcsin(\sin \alpha) = \arccos(\cos \alpha)$ .

Students often incorrectly assume this holds for all $\alpha \in [0, 2\pi]$ , leading them to believe the probability $p = P(X=1)$ can range up to $1$ .

If they fail to restrict $\alpha$ to the principal value branch $[0, \pi/2] \cup \{2\pi\}$ , they will incorrectly minimize the piecewise maximum function $M(p)$ at $p=2/3$ , yielding the trap answer $2$ .

The non-negativity restriction of the LP is standard, but the domain restriction on the objective coefficients is the true differentiator.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Linear Programming - Class 12
Concept	Non-Negativity Restriction
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free