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Question #41011

Difficulty: 8/10 | 🟢 Free

The Question

Let $M = \begin{bmatrix} 1 & -1 & -1 \\ 0 & 2 & 1 \\ 0 & 0 & 1 \end{bmatrix}$ . Let $\Omega$ be the region in $\mathbb{R}^3$ defined by the system of inequalities: $x \ge 0, y \ge 0, z \ge 0$ , $x + y + z \le 6$ , and $2x + y + z \le 13$ . A point $\mathbf{v} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is chosen uniformly at random from $\Omega$ . Let $E$ be the event that the vector $\mathbf{u} = \text{adj}(M) \mathbf{v}$ also lies in $\Omega$ . Find the probability of the event $E$ .

Options

A.
$\frac{1}{8}$
B.
$\frac{1}{16}$
C.
$\frac{1}{32}$
D.
$\frac{1}{64}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Algebra of Events|Adjoint-Determinant Relation|Redundant Constraints

Why This Question Is Tricky

Common Mistakes

The problem integrates 3D geometric probability with linear transformations via the adjoint matrix.

The primary trap is the presence of a 'Redundant Constraint' ( $2x+y+z \le 13$ ) which masks the true sample space $\Omega$ , a simple tetrahedron.

Students who fail to verify redundancy may attempt complex intersection calculations, leading to incorrect volumes.

A secondary trap lies in the transformation $\mathbf{u} = \text{adj}(M)\mathbf{v}$ , which introduces a hidden non-negativity constraint ( $y \ge z$ ).

Ignoring this constraint yields a probability of 1/8 instead of the correct 1/16, testing the student's rigor in applying all boundary conditions of the transformed space.

Quick Facts About This Question

Question Information

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