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Question #41021

Difficulty: 8/10 | 🟢 Free

The Question

Let $W$ be the $80^{\text{th}}$ word in the dictionary of all unique permutations of the letters $\text{A, A, B, C, D, E}$ . A fair 6-sided die has its faces marked with the letters $\text{A, A, B, C, D, E}$ . A 6-letter word $X$ is generated by rolling this die 6 times and recording the letters in the order they appear. What is the probability that $X$ is a permutation of the letters of $W$ , and has the same starting letter and the same ending letter as $W$ ?

Options

A.
$\frac{1}{486}$
B.
$\frac{1}{972}$
C.
$\frac{24}{15625}$
D.
$\frac{1}{1944}$

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

Equiprobability Fallacy|Rank of a Word|Cartesian Product Cardinality

Why This Question Is Tricky

Common Mistakes

This question tests the Equiprobability Fallacy and Cartesian Product Cardinality in the context of multiset permutations.

Students often correctly identify the 80th word (ADBAEC) and find there are 24 valid permutations.

However, when calculating the probability, the trap lies in assuming all $6^6$ possible outcomes of the die rolls correspond 1-to-1 with unique words, leading to the incorrect probability $24/6^6 = 1/1944$ .

They fail to recognize that words with repeated letters have a higher frequency of occurrence in the Cartesian product space.

Another trap is assuming the letters are chosen from the set of unique letters $\{A,B,C,D,E\}$ , leading to $24/5^6$ .

The correct approach requires calculating the probability of a specific sequence using the individual letter probabilities $P(A)=1/3$ and $P(\text{others})=1/6$ , and multiplying by the number of valid unique arrangements.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 11
Concept	Equiprobability Fallacy
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free