Problem 22 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(\mathrm{A}\) and \(\mathrm{B}\) are two events such that \(\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A} \cap \mathrm{B})\), then the incorrect statement amongst the following statements is: [Online April 9, 2014] (a) \(\mathrm{A}\) and \(\mathrm{B}\) are equally likely (b) \(\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{r}\right)=0\) (c) \(\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)=0\) (d) \(\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=1\)

Step-by-Step Solution

Answer

The incorrect statement is (a): \( \mathrm{A} \) and \( \mathrm{B} \) are equally likely.

Step 1: Understand the Given Condition

The given condition is that \( \mathrm{P}(\mathrm{A} \cup \mathrm{B}) = \mathrm{P}(\mathrm{A} \cap \mathrm{B}) \). According to the formula for the probability of the union of two events, \( \mathrm{P}(\mathrm{A} \cup \mathrm{B}) = \mathrm{P}(\mathrm{A}) + \mathrm{P}(\mathrm{B}) - \mathrm{P}(\mathrm{A} \cap \mathrm{B}) \). Equating this to \( \mathrm{P}(\mathrm{A} \cap \mathrm{B}) \), we have:

Step 2: Simplifying the Equation

\[ \mathrm{P}(\mathrm{A}) + \mathrm{P}(\mathrm{B}) - \mathrm{P}(\mathrm{A} \cap \mathrm{B}) = \mathrm{P}(\mathrm{A} \cap \mathrm{B}) \]Rearranging gives: \ \[ \mathrm{P}(\mathrm{A}) + \mathrm{P}(\mathrm{B}) = 2\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \]

Step 3: Evaluate Each Statement

- **(a)** "\( \mathrm{A} \) and \( \mathrm{B} \) are equally likely": This means \( \mathrm{P}(\mathrm{A}) = \mathrm{P}(\mathrm{B}) \). Without additional information, this isn't necessarily true from the equation, so it could be incorrect.- **(b)** "\( \mathrm{P}(\mathrm{A} \cap \mathrm{B}^r) = 0 \)": This implies \( \mathrm{A} \subseteq \mathrm{B} \), which agrees with our condition since outside of \( \mathrm{A} \cap \mathrm{B} \), no part of \( \mathrm{A} \) can occur.- **(c)** "\( \mathrm{P}(\mathrm{A}' \cap \mathrm{B}) = 0 \)": This implies \( \mathrm{B} \subseteq \mathrm{A} \), making \( \mathrm{B} \) completely covered by \( \mathrm{A} \), and aligns with the given condition.- **(d)** "\( \mathrm{P}(\mathrm{A}) + \mathrm{P}(\mathrm{B}) = 1 \)": This can be true if \( \mathrm{A} \) and \( \mathrm{B} \) combine to exactly cover the probability space when overlapping exactly as per the provided condition.

Step 4: Identify the Incorrect Statement

By the analysis in Step 3, the statement that does not hold based on the probability equation we derived and the possible condition setting is (a). Without explicit evidence or condition that the events \( \mathrm{A} \) and \( \mathrm{B} \) must have equal probabilities, this statement is not certain to be true.

Key Concepts

Union of events

In probability, the union of two events, denoted as \(\mathrm{A} \cup \mathrm{B}\), represents the occurrence of at least one of the events \(\mathrm{A}\) or \(\mathrm{B}\). Imagine you have two dice and want at least one to show a 'four'; this represents finding the probability of the union of two events. The formula used to compute this probability is:

• Probability of the union: \( \mathrm{P}(\mathrm{A} \cup \mathrm{B}) = \mathrm{P}(\mathrm{A}) + \mathrm{P}(\mathrm{B}) - \mathrm{P}(\mathrm{A} \cap \mathrm{B}) \)

You subtract \(\mathrm{P}(\mathrm{A} \cap \mathrm{B})\) to avoid double-counting the scenarios where both events occur. The union is crucial when we determine how events overlap and interact, providing a comprehensive view of event outcomes.

Intersection of events

The intersection of events, expressed as \(\mathrm{A} \cap \mathrm{B}\), refers to the event where both events \(\mathrm{A}\) and \(\mathrm{B}\) occur together. Picture a Venn diagram where the overlapping areas represent this intersection. Calculating the probability of this intersection is pivotal, especially when events depend on each other or there is a direct overlap.

• Probability of the intersection: It is directly provided by \( \mathrm{P}(\mathrm{A} \cap \mathrm{B}) \).

Understanding intersections aids in evaluating complex situations where outcomes are not isolated but occur concurrently. It helps in discerning combinations of events and their probabilities, leading to a better grasp of the underlying probability space.

Equally likely events

Equally likely events occur when two or more events have the same probability of happening. Think of flipping a fair coin, which results in heads or tails with equal chances. For two events \(\mathrm{A}\) and \(\mathrm{B}\) to be equally likely, both must satisfy:

• \(\mathrm{P}(\mathrm{A}) = \mathrm{P}(\mathrm{B})\)

In many textbook exercises, assumptions about events being equally likely help simplify problems. However, in our specific scenario \(\mathrm{P}(\mathrm{A} \cup \mathrm{B}) = \mathrm{P}(\mathrm{A} \cap \mathrm{B})\), equal likelihood isn't necessarily implied directly. Therefore, evaluating if events are equally likely requires checking beyond mere equal probability equations. Understanding this fosters critical thinking about conditions where we might incorrectly assume equal likelihood without proof.

Probability space

A probability space forms the framework within which probability problems are studied. It is analogous to the backdrop on which events unfold and consists of three main components:

• Sample Space (\(S\)): The set of all possible outcomes in an experiment.
• Events: Subsets of the sample space representing potential outcomes that might interest us, like obtaining a head in a coin toss.
• Probability Measure: A function assigning probabilities to events within the sample space, satisfying the probability axioms.

In the contest of \(\mathrm{P}(\mathrm{A} \cup \mathrm{B}) = \mathrm{P}(\mathrm{A} \cap \mathrm{B})\), the probability space must account for every possible scenario where the events interact. Understanding the layout of a probability space reveals whether certain conditions or constraints need to be met for statements about events and their probabilities to hold true. This conceptual approach simplifies the analysis and comparison of probability statements.