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Question #40551

Difficulty: 8/10 | 🟢 Free

The Question

A point $P$ is generated in $\mathbb{R}^3$ by the following random process: With probability $\frac{1}{3}$ , $P$ is chosen uniformly from the surface of the unit sphere $S = \{(x,y,z) : x^2+y^2+z^2 = 1\}$ . With probability $\frac{2}{3}$ , $P$ is chosen uniformly from the interior of the unit sphere $V = \{(x,y,z) : x^2+y^2+z^2 < 1\}$ . Let $O$ be the origin. Let $A$ be the event that the volume of the sphere centered at $O$ with radius $OP$ is at least $\frac{\pi}{6}$ . Let $B$ be the event that $P$ lies strictly inside $S$ . Find the probability $P(\overline{A} \cup \overline{B})$ .

Options

A.
$\frac{5}{12}$
B.
$\frac{1}{12}$
C.
$\frac{7}{12}$
D.
$\frac{11}{12}$

Check Your Answer

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

Closed vs Open Intervals|Axiomatic Probability Constraints|De Morgan's Laws in Probability

Why This Question Is Tricky

Common Mistakes

This question tests the 'Boundary Trap' in mixed probability distributions using De Morgan's Laws.

Students often incorrectly assume that continuous probability distributions have measure zero at all boundaries, ignoring the discrete probability mass at $X=1$ introduced by the piecewise definition.

By asking for $P(\overline{A} \cup \overline{B})$ , students must apply De Morgan's Laws to find $1 - P(A \cap B)$ or calculate $P(\overline{A}) + P(\overline{B})$ .

If they fail to strictly intersect the open interval $B = \{X < 1\}$ with $A$ , they will erroneously include the boundary point $X=1$ , leading to $P(A \cap B) = \frac{11}{12}$ and the trap answer $\frac{1}{12}$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Linear Inequalities - Class 11
Concept	Closed vs Open Intervals
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free