Problem 24 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(X\) and \(Y\) are two events such that \(P(X \cup Y)=P(X \cap Y)\) Statement 1: \(P\left(X \cap Y^{\prime}\right)=P\left(X^{\prime} \cap Y\right)=0\) Statement 2: \(P(X)+P(Y)=2 P(X \cap Y)\) [Online May 7, 2012] (a) Statement 1 is false, Statement 2 is true. (b) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1 . (c) Statement 1 is true, Statement 2 is false. (d) Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation of Statement 1 .

Step-by-Step Solution

Answer

(d) Statement 1 is true, Statement 2 is true; Statement 2 explains Statement 1.

Step 1: Understand the Given Condition

The problem states that \(P(X \cup Y) = P(X \cap Y)\). This means the probability of either event \(X\) or \(Y\) or both occurring is equal to the probability of both \(X\) AND \(Y\) occurring at the same time. This implies \(X\) and \(Y\) are dependent events.

Step 2: Evaluate Statement 1

Statement 1 claims that \(P(X \cap Y') = 0\) and \(P(X' \cap Y) = 0\). - \(X \cap Y'\) represents the event that \(X\) occurs but \(Y\) does not.- \(X' \cap Y \) represents the event that \(Y\) occurs but \(X\) does not.Since \(P(X \cup Y) = P(X \cap Y)\), it implies no probability is left for events where only one occurs, which means both these probabilities must be zero.

Step 3: Evaluate Statement 2

Statement 2 claims \(P(X) + P(Y) = 2P(X \cap Y)\).Using the condition \(P(X \cup Y) = P(X \cap Y)\), and the equation of probabilities, we have: \[P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)\]Replacing \(P(X \cup Y)\) with \(P(X \cap Y)\), we get:\[P(X \cap Y) = P(X) + P(Y) - P(X \cap Y)\]Simplifying gives:\[2P(X \cap Y) = P(X) + P(Y)\]This complexifies to confirm Statement 2 is true.

Step 4: Validate the Connection between Statements

Both Statement 1 and Statement 2 derive directly from the given condition \(P(X \cup Y) = P(X \cap Y)\). The condition implies \(P(X \cap Y')\) and \(P(X' \cap Y)\) must be zero. Additionally, the derived equation \(P(X) + P(Y) = 2P(X \cap Y)\) explicitly explains the probability distribution between \(X\) and \(Y\). Thus, Statement 2 not only supports but explains Statement 1.

Key Concepts

Dependent Events

Dependent events in probability are events where the outcome or occurrence of one event affects the outcome or occurrence of another. Understanding these relationships is crucial in probability, as it changes how we calculate the likelihood of combined or sequential events.

In the exercise, the given condition is that the probability of the union of two events, \(P(X \cup Y)\), is equal to the probability of their intersection, \(P(X \cap Y)\). This directly suggests that events \(X\) and \(Y\) are dependent. In such scenarios, if one event occurs, it invariably means the other must also occur because there's no leftover probability for a scenario where only one of them happens.

Recognizing events as dependent helps us anticipate outcomes and plan how to approach solving problems related to these events effectively.

Probability Distribution

Probability distribution involves understanding how probabilities are shared or distributed among different outcomes of an event. In the current exercise, it explores how the given condition affects the probability distribution between events \(X\) and \(Y\).

- **Joint Probability Distribution:** Here, \(P(X \cup Y) = P(X \cap Y)\) implies a unique probability distribution, where there is no probability left for outcomes where only \(X\) or only \(Y\) occurs.

Therefore, the entire probability is concentrated on the intersection of the two events, i.e., \(X \cap Y\). The implication is that neither \(X\) nor \(Y\) can occur independently, which shapes how we view the distribution of probabilities across these outcomes.

This type of probability setup is not common, hence analyzing such distributions can enhance our understanding of interdependent event relationships.

Probability Equations

Probability equations serve as mathematical representations that express relationships between different probabilities. They are critical in calculating outcomes and understanding complex interrelations in probabilistic scenarios.

In the exercise, one key probability equation derived is \(P(X \cap Y) = P(X) + P(Y) - P(X \cap Y)\). By simplifying it, we reach \(2P(X \cap Y) = P(X) + P(Y)\). This highlights how the equal distribution of probabilities across events \(X\) and \(Y\) is confirmed mathematically.

This equation helps validate Statement 2 in the exercise—showing the balance in probabilities—and emphasizes how dependent the events are on each other.

The power of probability equations lies in their ability to provide a concise mathematical framework that can accurately represent and predict the behavior of multiple events in probability theory.