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Question #41066

Difficulty: 8/10 | 🟢 Free

The Question

Let $\Omega$ be the set of all $6$ -letter words formed using the letters from the set $\{A, B, C, D\}$ . For $k \in \{1, 2, 3, 4\}$ , let $S_k$ be the subset of $\Omega$ consisting of words that contain exactly $k$ distinct letters. A subset $S_k$ is chosen at random with probability $\frac{1}{4}$ , and a word $W$ is drawn uniformly at random from it. Given that $W$ is a palindrome containing exactly two 'A's, what is the probability that $W$ belongs to $S_3$ ?

Options

A.
$\frac{31}{121}$
B.
$\frac{2}{3}$
C.
$\frac{31}{61}$
D.
$\frac{90}{121}$

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

Combinatorial Counting (P&C Integration)|Bayes' Theorem Reversal|The Empty Set Trap|Inclusion-Exclusion Principle

Why This Question Is Tricky

Common Mistakes

This problem tests the 'Empty Set Trap' within a Bayesian framework combined with the Inclusion-Exclusion Principle.

Students often blindly apply combinatorial formulas to find the intersection of events without verifying domain validity.

Here, forming a 6-letter palindrome with 4 distinct letters requires four positive even integers summing to 6, which is mathematically impossible.

Failing to recognize that the intersection with S_4 is an empty set leads to incorrect denominator terms in Bayes' theorem.

Furthermore, the problem distinguishes between a uniform draw from the entire sample space versus a two-step hierarchical selection, trapping students who ignore the varying sizes of the subsets when calculating conditional probabilities.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 11
Concept	Combinatorial Counting (P&C Integration)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free