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Question #41834

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the set of 8 points in $\mathbb{R}^3$ given by: $S = \{(0,0,0), (1,0,0), (2,0,0), (0,1,0), (0,2,0), (0,0,1), (0,0,2), (1,1,1)\}$ Three distinct points $A, B,$ and $C$ are chosen uniformly at random from $S$ . Given that the points $A, B,$ and $C$ are non-collinear (and thus uniquely determine a plane $P$ ), a fourth point $D$ is chosen uniformly at random from the remaining 5 points in $S$ . What is the probability that the scalar triple product $[\vec{AB} \ \vec{AC} \ \vec{AD}]$ is non-zero?

Options

A.
$\frac{208}{265}$
B.
$\frac{26}{35}$
C.
$\frac{193}{265}$
D.
$\frac{42}{53}$

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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

Combinatorial Probability|Scalar Triple Product Volume Logic|Distance from point to plane

Why This Question Is Tricky

Common Mistakes

This question tests combinatorial probability intertwined with 3D geometry, specifically focusing on the scalar triple product and coplanarity.

The core trap is 'coordinate degeneracy'—students often assume any 3 points define a plane and any 4 coplanar points contribute equally to the failure state.

However, if the base 3 points are collinear, the plane is undefined, which must be excluded from the conditional sample space.

Furthermore, when counting coplanar 4-tuples, students must carefully subtract cases where the chosen base triplet is collinear, leading to the subtle difference between $\binom{n}{3}$ and $\binom{n}{3}-k$ valid bases.

Failing to account for collinearity in either the denominator or the numerator leads to the trap options.

Quick Facts About This Question

Question Information

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