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Question #40499

Difficulty: 8/10 | 🟢 Free

The Question

Let $A$ and $B$ be two events in a probability space such that $P(A) = \frac{2}{3}$ and $P(B) = \frac{1}{2}$ . Consider the quadratic equation $z^2 - P(A \cup B)z + P(A \cap B) = 0$ , whose roots are $\alpha$ and $\beta$ . If $\alpha^3 + \beta^3 = \frac{143}{1728}$ , then the value of the conditional probability $P(A \mid A^c \cup B^c)$ is:

Options

A.
$\frac{5}{9}$
B.
$\frac{1}{2}$
C.
$\frac{2}{3}$
D.
$\frac{1}{3}$

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

Vieta’s Formulas in Complex Domain|Symmetry and Invariance|Independent vs Disjoint Events|Conditional Probability Foundations

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of complex polynomial roots and conditional probability, specifically targeting the false assumption of event independence or disjointness.

Students often bypass solving the cubic equation by incorrectly assuming $P(A \cap B) = P(A)P(B)$ (independence) or $P(A \cap B) = 0$ (disjoint), which leads directly to the trap options 1/2 and 2/3 respectively.

The problem forces the use of Vieta's formulas and the probability addition theorem to form a cubic in $P(A \cup B)$ , rewarding candidates who rely on algebraic invariance rather than unwarranted probabilistic assumptions.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Complex Numbers and Quadratic Equations - Class 11
Concept	Vieta’s Formulas in Complex Domain
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free