Let H₁, H₂, ..., Hₙ be mutually exclusive and exhaustive events with P(Hᵢ) 0, i = 1, 2, ..., n. Le

Q: Wait, why do we reject the inequality? Isn't $P(E)$ always less than 1, so $\frac{1}{P(E)}$ is greater than 1, making the left side bigger?

That logic is sound *only* when the intersection is non-zero! If $P(H_i \cap E) = 0$, the inequality becomes $0 > 0$, which is false. In mathematics, a statement must hold for *all* cases. One counter-example is enough to break the entire claim.

Q: Is Statement-2 just a definition?

Yes, it is the definition of exhaustive events. Since the union of $H_i$ is the sample space $S$, the sum of their probabilities must be $P(S) = 1$. It is a fundamental axiom of probability theory.

Q: Why does the problem specify $0 < P(E) < 1$?

This ensures that $P(E)$ is neither 0 (which would make conditional probability undefined) nor 1 (which would make the inequality trivial). It sets the stage for the 'scaling' effect we discussed, forcing us to think about the relationship between the intersection and the total probability.

Question

Let $H_1, H_2, \dots, H_n$ be mutually exclusive and exhaustive events with $P(H_i) > 0, i = 1, 2, \dots, n$. Let $E$ be any other event with $0 < P(E) < 1$. STATEMENT-1: $P(H_i|E) > P(E|H_i) \cdot P(H_i)$ for $i = 1, 2, \dots, n$ because STATEMENT-2: $\sum_{i=1}^n P(H_i) = 1$. Options: (A) Statement-1 is True, statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.

Sigma Percentile · Accepted Answer

### The Architecture of Probability: A Journey Through Partitions Welcome, future engineer. Today, we are not just solving a probability problem; we are dissecting the very structure of how we view uncertainty. When we talk about a sample space $S$ partitioned into events $H_1, H_2, \dots, H_n$, we are building a map. Imagine a vast, flat landscape where these events are the provinces that divide this land. Because they are mutually exclusive, no two provinces overlap. Because they are exhaustive, they cover the entire map without leaving a single patch of dirt unclaimed. This is the bedrock of our problem. ### Phase 1: The Truth of Statement-2 Let us look at Statement-2: $\sum_{i=1}^n P(H_i) = 1$. Is this true? Absolutely. Since these events are exhaustive, their union is the entire sample space $S$. By the fundamental axioms of probability, the probability of the entire sample space is $P(S) = 1$. Therefore, the sum of the probabilities of these disjoint, exhaustive events must equal $1$. This is a rock-solid truth and our first anchor point. ### Phase 2: The Conditional Lens Now, we introduce event $E$. Think of $E$ as a spotlight shining onto our map. It might illuminate parts of $H_1$, parts of $H_2$, or perhaps it misses some provinces entirely. We are asked to evaluate the inequality $P(H_i|E) > P(E|H_i) \cdot P(H_i)$. To understand this, we must use the definition of conditional probability. We know that $P(H_i|E) = \frac{P(H_i \cap E)}{P(E)}$. This is the probability of $H_i$ occurring, given that we are restricted to the region illuminated by $E$. Simultaneously, the multiplication rule tells us that $P(H_i \cap E) = P(E|H_i)P(H_i)$. If we substitute this into our conditional probability formula, we get: $$P(H_i|E) = \frac{P(E|H_i)P(H_i)}{P(E)}$$ ### Phase 3: The Algebraic Trap Here is where the intuition kicks in. We are given $0 < P(E) < 1$. This is a crucial piece of information. If $P(E)$ is a fraction between $0$ and $1$, then its reciprocal, $\frac{1}{P(E)}$, is strictly greater than $1$. Let us rewrite our equation to highlight this: $$P(H_i|E) = \frac{1}{P(E)} \cdot [P(E|H_i)P(H_i)]$$ If we look at this, it seems like we are multiplying the term $P(E|H_i)P(H_i)$ by a factor greater than $1$. If the intersection $P(H_i \cap E)$ is positive, then yes, $P(H_i|E)$ will indeed be strictly greater than $P(E|H_i)P(H_i)$. It feels like we have proven the inequality. ### Phase 4: The Moment of Clarity But wait. In JEE Advanced, the devil is always in the details. The statement claims this inequality holds for *all* $i = 1, 2, \dots, n$. What if there exists an event $H_k$ such that $H_k \cap E = \emptyset$? In this scenario, the intersection is empty, and the probability $P(H_k \cap E) = 0$. If we plug this into our inequality, we get $0 > 0$. This is mathematically false. The inequality fails. Because the problem does not guarantee that $E$ intersects with every single partition $H_i$, we cannot claim the inequality holds for all $i$. ### Conclusion This is the beauty of the problem. It teaches us that in mathematics, a universal quantifier like 'for all' is a heavy burden. It only takes one counter-example—one province that the spotlight of $E$ misses—to bring the entire statement crashing down. Statement-2 is true, but Statement-1 is false. We have navigated the trap, understood the geometry of the sample space, and emerged with the correct answer: **Option (d)**. Keep this rigor in your toolkit; it will serve you well.

Select Answer:

Visualized Solution

Visualizing the Sample Space

Analyzing Statement 2

Introducing Event $E$

Defining Conditional Probability

Applying the Multiplication Rule

Analyzing the Magnitude of $P (E)$

Testing the Inequality (The Positive Case)

The Critical Counter-example (The Trap)

Final Verdict

The Sigma Insight: Bayes' Theorem

The Architecture of Probability

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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Select Answer:

Visualized Solution

Visualizing the Sample Space

Analyzing Statement 2

Introducing Event .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

Defining Conditional Probability

Applying the Multiplication Rule

Analyzing the Magnitude of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E)

Testing the Inequality (The Positive Case)

The Critical Counter-example (The Trap)

Final Verdict

The Sigma Insight: Bayes' Theorem

The Architecture of Probability

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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