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Question #41008

Difficulty: 8/10 | 🟢 Free

The Question

In a geometric probability model, the probability of an event $E$ is given by the formula $P(E) = \frac{1}{10} [\vec{u}\ \vec{v}\ \vec{w}]$ , where $[\vec{u}\ \vec{v}\ \vec{w}]$ denotes the scalar triple product $\vec{u} \cdot (\vec{v} \times \vec{w})$ . Let $\vec{u} = x\hat{i} + 3\hat{j} - \hat{k}$ , $\vec{v} = \hat{i} - 2\hat{j} + \hat{k}$ , and $\vec{w} = 2\hat{i} + x\hat{j} + 2\hat{k}$ . If $x$ is an integer, what is the sum of all possible values of $x$ for which $P(E)$ satisfies the axioms of probability?

Options

A.
-10
B.
-20
C.
-5
D.
0

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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

Axiomatic Probability Constraints|Scalar Triple Product Volume Logic|Dot Product Projection Sign

Why This Question Is Tricky

Common Mistakes

This question tests the fundamental Kolmogorov axioms of probability, specifically the non-negativity constraint $P(E) \ge 0$ , integrated with vector geometry.

A common trap for JEE Advanced students is to instinctively apply the absolute value to the scalar triple product because it represents geometric volume.

If a student incorrectly assumes $P(E) = \frac{|[\vec{u}\ \vec{v}\ \vec{w}]|}{10}$ , they will solve $|-x^2 - 5x - 4| \le 10$ , leading to the erroneous interval $x \in [-6, 1]$ and a sum of $-20$ .

The problem strictly defines $P(E)$ without the modulus, forcing the student to use the probability axioms to reject negative scalar triple products.

Quick Facts About This Question

Question Information

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