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Let \(\mathrm{A}\) and \(\mathrm{E}\) be any two events with positive probabilities: Statement \(-1: \mathrm{P}(\mathrm{E} / \mathrm{A}) \geq \mathrm{P}(\mathrm{A} / \mathrm{E}) \mathrm{P}(\mathrm{E})\) Statement - 2: \(\mathrm{P}(\mathrm{A} / \mathrm{E}) \geq \mathrm{P}(\mathrm{A} \cap \mathrm{E})\) \mathrm{\\{} O n l i n e ~ A p r i l ~ 1 9 , ~ 2 0 1 4 ] ~ (a) Both the statements are true (b) Both the statements are false (c) Statement- 1 is true, Statement- 2 is false (d) Statement- 1 is false, Statement- 2 is true

Short Answer

Expert verified
(d) Statement 1 is false, Statement 2 is true.

Step by step solution

01

Understanding Statement 1

The first statement is \( \mathrm{P}(\mathrm{E} / \mathrm{A}) \geq \mathrm{P}(\mathrm{A} / \mathrm{E}) \cdot \mathrm{P}(\mathrm{E}) \). By the definition of conditional probability, \( \mathrm{P}(\mathrm{E} / \mathrm{A}) = \frac{\mathrm{P}(\mathrm{E} \cap \mathrm{A})}{\mathrm{P}(\mathrm{A})} \) and \( \mathrm{P}(\mathrm{A} / \mathrm{E}) = \frac{\mathrm{P}(\mathrm{A} \cap \mathrm{E})}{\mathrm{P}(\mathrm{E})} \). Simplifying, \( \mathrm{P}(\mathrm{A} \cap \mathrm{E}) = \mathrm{P}(\mathrm{E} / \mathrm{A}) \cdot \mathrm{P}(\mathrm{A}) \) and \( \mathrm{P}(\mathrm{A} \cap \mathrm{E}) = \mathrm{P}(\mathrm{A} / \mathrm{E}) \cdot \mathrm{P}(\mathrm{E}) \). Thus, the statement suggests \( \frac{\mathrm{P}(\mathrm{E} \cap \mathrm{A})}{\mathrm{P}(\mathrm{A})} \geq \frac{\mathrm{P}(\mathrm{A} \cap \mathrm{E})}{\mathrm{P}(\mathrm{A})} \). This simplifies to a correct equality, \( \mathrm{P}(\mathrm{A} \cap \mathrm{E}) \geq \mathrm{P}(\mathrm{A} \cap \mathrm{E}) \cdot \frac{\mathrm{P}(\mathrm{E})}{\mathrm{P}(\mathrm{A})} \), indicating the statement is false.
02

Understanding Statement 2

The second statement is \( \mathrm{P}(\mathrm{A} / \mathrm{E}) \geq \mathrm{P}(\mathrm{A} \cap \mathrm{E}) \). We know \( \mathrm{P}(\mathrm{A} / \mathrm{E}) = \frac{\mathrm{P}(\mathrm{A} \cap \mathrm{E})}{\mathrm{P}(\mathrm{E})} \). The inequality \( \mathrm{P}(\mathrm{A} \cap \mathrm{E}) \leq \frac{\mathrm{P}(\mathrm{A} \cap \mathrm{E})}{\mathrm{P}(\mathrm{E})} \) would imply \( \mathrm{P}(\mathrm{E}) \leq 1 \). Since \( \mathrm{P}(\mathrm{E}) \leq 1 \) is always true for a probability, this statement holds true. Therefore, Statement 2 is true.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability of Events
Understanding the probability of events is fundamental to mastering probability theory. When we talk about probability, we essentially mean the measure of how likely it is for an event to happen. For any event \( A \), the probability \( \mathrm{P}(A) \) must lie between 0 and 1. 0 indicates that the event will not occur, while 1 means the event is certain to occur. An important property of probabilities is that the sum of all possible outcomes must be equal to 1. Therefore, when you calculate \( \mathrm{P}(A) \), you're calculating the chance that \( A \) occurs out of all possible outcomes.

In problems with more than one event, like \( A \) and \( E \) in our exercise, conditional probability often comes into play. Conditional probability \( \mathrm{P}(A / E) \) is defined as the probability that event \( A \) occurs given that \( E \) has already occurred. This can be calculated using the formula \( \mathrm{P}(A / E) = \frac{\mathrm{P}(A \cap E)}{\mathrm{P}(E)} \), where \( A \cap E \) represents the probability that both A and E happen together. Understanding these foundational concepts of probability sets the stage for tackling more complex mathematical concepts and problems.
Inequalities in Probability
In the realm of probability, inequalities can help us establish relations between different conditional probabilities. Let's break down the kind of inequality presented in the original problem: \( \mathrm{P}(E / A) \geq \mathrm{P}(A / E) \cdot \mathrm{P}(E) \). Evaluating conditional probabilities often means testing such inequalities to determine their validity.

For any event with a conditional probability, these derived inequalities can tell us valuable information about relationships between events but must be approached carefully. Interface between conditional probabilities can occasionally produce false inequalities if not thoroughly analyzed.

Checking for the validity of inequalities can include simplifying each part using known formulas. In our exercise, the exercise showed an initial inequality getting simplified to reveal an incorrect relation, meaning the inequality does not hold true. Being attentive to the principles of probability helps us identify these logical fallacies. Always remember, probability values remain bounded between 0 and 1, and any inequality involving them should respect these constraints.
JEE Main Mathematics Solutions
To solve problems from competitive exams like JEE Main, a good grasp of the core mathematical concepts, including probability, is essential. The topics covered in JEE Main mathematics often involve integrating knowledge across various mathematical domains, including inequalities and probability.

In the context of the given problem, understanding how conditional probabilities and inequalities operate can significantly boost a student's ability to discern between true or false statements. By applying structured logical reasoning, often involved in detailed step-by-step solutions, students can tackle even the most confusing seeming questions.

Practicing such problems helps reinforce the theoretical understanding necessary for tackling similar questions in an exam setting. Analyzing statements, like the two given in the exercise above, not only shows theoretical knowledge but also the ability to critically evaluate mathematical relationships. Ultimately, detailed solutions like those found in JEE Main mathematics prepare students not only for exams but for higher-level mathematical thinking and application.

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Most popular questions from this chapter

Let \(A, B, C\), be pairwise independent events with \(P(\mathrm{C})>0\) and \(P(A \cap B \cap C)=0\). Then \(P\left(A^{c} \cap B^{c} / C\right)\). [2011RS] (a) \(\mathrm{P}\left(B^{c}\right)-\mathrm{P}(B)\) (b) \(P\left(A^{c}\right)+P\left(B^{c}\right)\) (c) \(P\left(A^{c}\right)-P\left(B^{c}\right)\) (d) \(P\left(A^{c}\right)-P(B)\)

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