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Question #41720

Difficulty: 8/10 | 🟢 Free

The Question

A point $P(x,y,z)$ is chosen uniformly at random from the solid sphere $x^2 + y^2 + z^2 \le 9$ . Let $D$ be the perpendicular distance from $P$ to the plane $\Pi: x + 2y + 2z = 3$ . Two boxes, $B_1$ and $B_2$ , contain colored balls. Box $B_1$ contains 2 white and 3 black balls. Box $B_2$ contains 4 white and 1 black balls. If $D < 1$ , Box $B_1$ is selected. If $D \ge 1$ , Box $B_2$ is selected. A ball is drawn at random from the selected box and is found to be white. The probability that Box $B_1$ was selected is:

Options

A.
$\frac{23}{85}$
B.
$\frac{25}{29}$
C.
$\frac{13}{41}$
D.
$\frac{17}{27}$

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

Bayes' Theorem Reversal|Line-Plane intersection logic|Distance from point to plane

Why This Question Is Tricky

Common Mistakes

This question tests the integration of 3D coordinate geometry with Bayes' Theorem, specifically targeting the handling of absolute value constraints in distance formulas.

The core trap lies in the definition of the partition $D < 1$ .

Students frequently drop the absolute value in the perpendicular distance formula, incorrectly solving $\frac{x+2y+2z-3}{3} < 1 \implies x+2y+2z < 6$ .

This leads them to calculate the volume of the sphere for $u < 2$ (yielding $P(B_1) = 25/27$ ) instead of the correct bounded region $0 < u < 2$ (yielding $P(B_1) = 23/54$ ).

This single sign error in the geometric phase propagates through the posterior probability calculation, leading directly to the trap option $25/29$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 12
Concept	Bayes' Theorem Reversal
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free