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Question #41824

Difficulty: 8/10 | 🟢 Free

The Question

Let $X$ and $Y$ be independent random variables, each taking values in the set $S = \{1, 2, 3, \dots, 9\}$ . The probability mass function for both $X$ and $Y$ is given by $P(X=k) = P(Y=k) = \frac{k}{45}$ for all $k \in S$ . Given that $X + Y$ is a multiple of 3, what is the conditional probability that $X = Y$ ?

Options

A.
$\frac{7}{38}$
B.
$\frac{1}{9}$
C.
$\frac{14}{75}$
D.
$\frac{14}{225}$

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

Probability Mass Function (PMF)|Conditional Probability Constraints|Equivalence Class Partitioning

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to apply conditional probability over non-uniform probability mass functions partitioned by equivalence classes.

The primary trap is the 'uniformity assumption'—students often identify the 27 valid pairs $(x,y)$ that sum to a multiple of 3 and note that 3 of these pairs satisfy $x=y$ , incorrectly concluding the probability is $3/27 = 1/9$ .

This bypasses the PMF weights entirely.

Another trap is failing to normalize by the true probability of the condition $P(A)$ , or assuming $P(A) = 1/3$ as it would be in a continuous uniform distribution, leading to incorrect scaling of the joint probability.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 12
Concept	Probability Mass Function (PMF)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free