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Question #41077

Difficulty: 8/10 | 🟢 Free

The Question

A point $P(x,y,z)$ is chosen at random from the solid sphere $x^2+y^2+z^2 \le 1$ . The probability density function of the chosen point is proportional to its squared distance from the origin, i.e., $f(x,y,z) = k(x^2+y^2+z^2)$ for some constant $k$ . Let $E$ be the event that the perpendicular distance from $P$ to the plane $2x - y + 2z = 0$ is at most $\frac{1}{2}$ . Let $F$ be the event that $2x - y + 2z \ge 0$ . Find the conditional probability $P(E|F)$ .

Options

A.
$\frac{79}{128}$
B.
$\frac{11}{16}$
C.
$\frac{6479}{31104}$
D.
$\frac{107}{432}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Conditional Probability Foundations|Geometry Constraint Modeling|Distance from point to plane|Unit Vector Normalization

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to model geometric probability with non-uniform density and correctly apply distance constraints in 3D space.

The primary 'Monster Trap' lies in the unit vector normalization: students frequently compute the distance to the plane $2x-y+2z=0$ simply as $|2x-y+2z|$ , forgetting to divide by the normal vector's magnitude ( $\sqrt{2^2+(-1)^2+2^2} = 3$ ).

This leads to an incorrect integration bound ( $u \le 1/6$ instead of $u \le 1/2$ ).

Furthermore, the problem demands a rigorous change of coordinates to integrate the radial density function over planar slices, effectively filtering out candidates who rely on rote volume formulas.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 11
Concept	Conditional Probability Foundations
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free