Start Learning Free

Before You Begin

Question #41715

Difficulty: 8/10 | 🟢 Free

The Question

Let AA and BB be two independent events with probabilities P(A)=12P(A) = \frac{1}{2} and P(B)=qP(B) = q, where q∈(0,1)q \in (0, 1). Consider the matrix M=(P(A∪B)P(Ac∩Bc)−12P(A∩B)P(Ac∪Bc)−12P(A)P(Ac)−12)M = \begin{pmatrix} P(A \cup B) & P(A^c \cap B^c) & -\frac{1}{2} \\ P(A \cap B) & P(A^c \cup B^c) & -\frac{1}{2} \\ P(A) & P(A^c) & -\frac{1}{2} \end{pmatrix} Let SS be the set of all values of qq for which the eigenvectors of MM span R3\mathbb{R}^3. Then the sum of all elements in SS is:

Options

  • A.

    12\frac{1}{2}

  • B.

    1

  • C.

    14\frac{1}{4}

  • D.

    S \text{ is an empty set}

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

Independent vs Disjoint Events|Linear Independence vs Dependence|Even Multiplicity Roots

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of probability and linear algebra, specifically targeting the 'Even Multiplicity' trap.

Students often assume that if a matrix is formed by continuous or distinct probabilistic parameters, its eigenvalues will be distinct, guaranteeing diagonalizability.

Here, the characteristic equation yields a repeated root (λ=1/2\lambda = 1/2) regardless of the value of qq.

The trap forces students to manually check the nullity of (M−λI)(M - \lambda I) to find the geometric multiplicity.

If they blindly assume diagonalizability due to the variable qq, or miscalculate the span of the eigenspace, they will incorrectly conclude that SS contains all q∈(0,1)q \in (0,1) or is empty.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterProbability - Class 12
ConceptIndependent vs Disjoint Events
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree