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Question #40614

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be a set of 12 points in a plane, consisting of the 4 vertices of a square and 8 additional points. These 8 points are placed such that exactly 2 points lie strictly on each of the 4 line segments forming the sides of the square. Assume no other set of 3 or more points from $S$ are collinear. If a subset of 5 points is chosen at random from $S$ , the probability that no 3 of the chosen points are collinear is:

Options

A.
$\frac{29}{66}$
B.
$\frac{13}{33}$
C.
$\frac{47}{99}$
D.
$\frac{23}{66}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Geometry-based Combinations|Combinatorial Probability|Cardinality Inclusion-Exclusion

Why This Question Is Tricky

Common Mistakes

This question tests the application of the Inclusion-Exclusion Principle (IEP) in combinatorial geometry.

The primary trap is the over-counting of subsets that contain collinear points from multiple intersecting lines.

Students often calculate the total illegal combinations as simply the sum of illegal subsets on each line ( $\sum |A_i| = 480$ ), forgetting that a 5-point subset can simultaneously contain 3 points from one side and 3 points from an adjacent side if it includes their shared vertex.

Failing to subtract this intersection ( $\sum |A_i \cap A_j| = 36$ ) leads to the classic double-counting error and the incorrect answer of 13/33.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Permutations and Combinations - Class 11
Concept	Geometry-based Combinations
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free