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Question #41044

Difficulty: 8/10 | 🟢 Free

The Question

Let $X$ be a continuous random variable with the probability density function: $f(x) = \begin{cases} \frac{1}{2} \left(1 + \sin\left(\frac{\pi x}{2}\right)\right), & x \in [-1, 1] \\ 0, & \text{otherwise} \end{cases}$ Let $A$ and $B$ be events defined by $A = \{X > 0\}$ and $B = \{X < -1/3\}$ . Calculate the probability $P(A^c \cap B^c)$ .

Options

A.
$\frac{1}{6} + \frac{\sqrt{3} - 2}{2\pi}$
B.
$\frac{1}{6} - \frac{\sqrt{3} + 2}{2\pi}$
C.
$\frac{1}{6} + \frac{\sqrt{3} + 2}{2\pi}$
D.
$\frac{1}{6} + \frac{2 - \sqrt{3}}{2\pi}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

De Morgan’s Laws in Probability|Even and Odd Function Symmetry|Symmetry Utilization

Why This Question Is Tricky

Common Mistakes

The question tests the application of De Morgan's Laws to continuous probability distributions and the careful evaluation of integrals involving odd functions.

The primary trap occurs when students use $P(A^c \cap B^c) = 1 - P(A) - P(B)$ and attempt to simplify the evaluation of $P(B) = P([-1, -1/3))$ by exploiting the symmetric domain $[-1, 1]$ .

Assuming $P([-1, -1/3)) = P([1/3, 1])$ is a fatal error because the PDF contains an odd component ( $\sin(\frac{\pi x}{2})$ ) which shifts signum across the y-axis.

This false symmetry shortcut leads directly to the incorrect option B, penalizing students who fail to verify if the event space behaves symmetrically under the complement operation.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 11
Concept	De Morgan’s Laws in Probability
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free