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Question #41043

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be a set of 10 points in 3D space consisting of the 4 vertices and the 6 edge midpoints of a regular tetrahedron. Four points are chosen at random from $S$ . Let $A$ be the event that the 4 chosen points are coplanar, and $B$ be the event that no three of the chosen points are collinear. Find the probability of the event $A^c \cup B^c$ .

Options

A.
$\frac{61}{70}$
B.
$\frac{47}{70}$
C.
$\frac{31}{35}$
D.
$\frac{9}{70}$

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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

De Morgan’s Laws in Probability|Coplanarity Condition|Collinearity vs Coplanarity

Why This Question Is Tricky

Common Mistakes

This question tests the application of De Morgan's Laws in a 3D geometric probability context.

The primary trap lies in the geometric counting: students often assume that any coplanar set of 4 points must contain a collinear triplet (an edge), leading them to incorrectly conclude $A \cap B = \emptyset$ .

Furthermore, even if they analyze the faces correctly, they frequently miss the 'hidden' coplanar sets formed by the 3 equatorial squares (parallelograms of midpoints).

By asking for $P(A^c \cup B^c)$ , the problem forces the use of $1 - P(A \cap B)$ , punishing those who attempt to calculate the complex union directly or misunderstand the subtle distinction between coplanarity and collinearity in a tetrahedral grid.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 11
Concept	De Morgan’s Laws in Probability
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free