Problem 128 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(A\) and \(B\) are two events such that \(P(A)=\frac{1}{2}\) and \(P(B)=\frac{2}{3}\), then (A) \(P(A \cup B) \geq \frac{2}{3}\) (B) \(P\left(A \cap B^{\prime}\right) \leq \frac{1}{3}\) (C) \(1 / 6 \leq P(A \cap B) \leq \frac{1}{2}\) (D) \(1 / 6 \leq P\left(A^{\prime} \cap B\right) \leq \frac{1}{2}\)

Step-by-Step Solution

Answer

Options (C) and (D) are valid based on logical examination.

Step 1: Understand Set Operations

To solve the problem, we need to understand the operations involved with events. The union of two events, denoted as \(A \cup B\), refers to either event \(A\) or event \(B\) or both occurring. The intersection, denoted \(A \cap B\), refers to both events \(A\) and \(B\) occurring simultaneously. The complement, denoted as \(A^{\prime}\), is the event that \(A\) does not occur.

Step 2: Use Probability Laws

Key probability rules include: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\), \(P(A^{\prime}) = 1 - P(A)\), and \(P(A \cap B^{\prime}) = P(A) - P(A \cap B)\). These formulas will help determine the probability of the various expressions given in the problem.

Step 3: Analyze Each Option

We analyze each option using the probability laws.1. **Option A: \(P(A \cup B) \geq \frac{2}{3}\)** Using the union rule: \(P(A \cup B) = \frac{1}{2} + \frac{2}{3} - P(A \cap B)\). Simplifying gives \(P(A \cup B) = \frac{7}{6} - P(A \cap B)\). Since probabilities cannot exceed 1, we need \(P(A \cap B) \leq \frac{1}{6}\).2. **Option B: \(P(A \cap B^{\prime}) \leq \frac{1}{3}\)** Apply the formula: \(P(A \cap B^{\prime}) = \frac{1}{2} - P(A \cap B)\). Since \(P(A \cap B) \geq 0\), this will always be \(\leq \frac{1}{2}\), so \(P(A \cap B^{\prime}) \leq \frac{1}{3}\) is possible if \(P(A \cap B) \geq \frac{1}{6}\).3. **Option C: \(\frac{1}{6} \leq P(A \cap B) \leq \frac{1}{2}\)** Refer to the union formula to find possible values for \(P(A \cap B)\) subject to conditions derived above. It satisfies this range when considering the constraints.4. **Option D: \(\frac{1}{6} \leq P(A^{\prime} \cap B) \leq \frac{1}{2}\)** Compute \(P(A^{\prime} \cap B) = \frac{2}{3} - P(A \cap B)\). Depending on \(P(A \cap B)\), this inequality is valid if \(\frac{1}{6} \leq P(A \cap B)\leq \frac{1}{2}\).

Step 4: Conclusion and Selection

After analyzing the inequalities and the inclusion/exclusion formulas for involving events, it's clear that the range statements (C) and (D) align with possible values of \(P(A \cap B)\) and \(P(A^{\prime} \cap B)\). Hence these options are correctly tailored by the rules, while A and B are not confirmed as strictly valid under all conditions.

Key Concepts

Set Operations

Set operations are fundamental to understanding probabilities involving multiple events. In probability theory, events are typically represented as sets. The basic operations you might encounter with these sets include the union, intersection, and complement. These operations allow us to describe complex events in simple mathematical terms.

• Union: The union of two sets or events, denoted as \(A \cup B\), includes all outcomes that are in either set \(A\), set \(B\), or in both. If you imagine two circles overlapping each other on a Venn diagram, the union would be anything that falls into either circle.
• Intersection: The intersection, represented by \(A \cap B\), contains only the elements common to both sets \(A\) and \(B\). Using our Venn diagram analogy, this would be the area where the two circles overlap.
• Complement: The complement of an event, denoted by \(A^{\prime}\), consists of all outcomes in the sample space that are not in set \(A\). It's like everything outside circle \(A\) in your Venn diagram.

Understanding these operations helps in solving probability problems as they form the basis of rules and laws of probability, which dictate how these events behave under different scenarios.

Union of Events

The union of events is a critical concept in probability, especially when determining the likelihood of multiple events occurring. When we talk about the union of events \(A \cup B\), we're essentially asking: "What is the probability that either event \(A\), event \(B\), or both happen?"
The formula used to find this probability is:\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]This formula takes into account the overlap between the sets \(A\) and \(B\) by subtracting \(P(A \cap B)\), since it is counted twice if both occurrences are added separately. This principle is based on the inclusion-exclusion formula which ensures we don't over-count the joint occurrences.
In everyday terms, think of this as calculating the probability of having either coffee or tea or both available at a party, without counting people who drink both more than once.
Understanding the union of events and applying the formula correctly helps make accurate predictions in probability, ensuring comprehensive coverage of all possible outcomes.

Complement of an Event

The complement of an event provides a way to understand the likelihood of an event not occurring. When we denote an event as \(A^{\prime}\), we describe the set of outcomes where event \(A\) does not happen. The probability of the complement of \(A\) is calculated as:
\[P(A^{\prime}) = 1 - P(A)\]This formula works because the probabilities of all possible outcomes in a sample space must sum to 1. Therefore, the probability that event \(A\) does not occur is just 1 minus the probability that it does occur.
For instance, if there's a 60% chance of rain today, the complement tells us there's a 40% chance of no rain, which might help you decide whether to carry an umbrella.
This concept is particularly useful for calculating complex probabilities by focusing on what’s left when an event does not happen. Understanding complements helps in a wide array of statistical and probability problems, allowing for a more holistic view of scenario outcomes.