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Question #41057

Difficulty: 8/10 | 🟢 Free

The Question

A point $(X,Y)$ is chosen uniformly at random from the region $D = \{(x,y) \in \mathbb{R}^2 : x^2 \le y \le 1\}$ . Let $A$ be the event that $|X| \le \frac{1}{2}$ and $B$ be the event that $Y \le \frac{1}{2}$ . The value of $P(A \cup B)$ is:

Options

A.
$\frac{3 + 2\sqrt{2}}{8}$
B.
$\frac{11}{16}$
C.
$\frac{44 + 5\sqrt{2}}{64}$
D.
$\frac{11 + 4\sqrt{2}}{16}$

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

Mutually Exclusive vs Independent|Symmetry Utilization|Horizontal vs Vertical Strips

Why This Question Is Tricky

Common Mistakes

This question tests the integration of a joint PDF over a non-rectangular support, specifically targeting the 'Horizontal vs Vertical Strips' trap and the false assumption of independence.

$Students often calculate P(A \cap B) using horizontal strips but fail to split the integral at y=1/4 (where the boundary changes from x=\pm\sqrt{y} to x=\pm 1/2), erroneously yielding P(A \cap B) = P(B) and leading to the trap option B.$

Alternatively, treating the variables as independent due to the rectangular nature of the events (ignoring the dependent support D) leads to trap option C.

Using vertical strips elegantly bypasses the boundary split, rewarding strategic symmetry utilization and Fubini's Theorem application.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 11
Concept	Mutually Exclusive vs Independent
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free