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Question #40298

Difficulty: 8/10 | 🟢 Free

The Question

Let $S = \{1, 2, 3, \dots, 10\}$ . A subset $A$ of $S$ is chosen uniformly at random from the set of all subsets of $S$ that do not contain any two consecutive integers. Let $E$ be the event that $A$ is a proper subset of $\{1, 3, 5, 7, 9\}$ , and $F$ be the event that $5 \in A$ . The conditional probability $P(E \mid F)$ is:

Options

A.
$\frac{3}{8}$
B.
$\frac{2}{5}$
C.
$\frac{5}{48}$
D.
$\frac{15}{512}$

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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

Subset vs. Proper Subset|Conditional Probability Foundations|Gap Method (Non-adjacency)

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of restricted sample spaces (Gap Method) and conditional probability, specifically targeting the 'Proper Subset' definition.

The primary trap lies in the denominator calculation: students often default to $2^{n-1}$ for subsets containing a specific element, ignoring the non-adjacency constraint which requires the Fibonacci sequence (or manual counting of valid gaps) to find $|F| = 40$ .

A secondary lethal trap is the 'Proper Subset' condition in the numerator; students who correctly navigate the Gap Method often forget to subtract 1 from the $2^4$ valid odd subsets, leading to the incorrect option 2/5.

The collision of these two constraints ensures only students with robust structural understanding succeed.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Sets - Class 11
Concept	Subset vs. Proper Subset
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free