Problem 131 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(P(A)=\frac{2}{5}\) and \(P(B)=\frac{4}{5}\), then (A) \(P(A \cup B) \geq \frac{4}{5}\) (B) \(\frac{1}{5} \leq P(A \cap B) \leq \frac{2}{5}\) (C) \(\frac{1}{4} \leq P(A / B) \leq \frac{1}{2}\) (D) \(P\left(A \cap B^{\prime}\right) \leq \frac{1}{5}\)

Step-by-Step Solution

Answer

Options (B) and (C) are correct, others are not always true.

Step 1: Understand the Given Probabilities and Terms

We are given probabilities of two events, \(A\) and \(B\), such that \(P(A) = \frac{2}{5}\) and \(P(B) = \frac{4}{5}\). We need to examine the inequalities regarding these probabilities and their combinations, including union, intersection, conditional probability, and complements.

Step 2: Evaluate \(P(A \cup B)\)

According to the addition rule for probabilities, \[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]. Since \(P(A) = \frac{2}{5}\) and \(P(B) = \frac{4}{5}\), the minimum value for \(P(A \cap B)\) is zero. Substituting these values, \(P(A \cup B) \leq P(A) + P(B) = \frac{2}{5} + \frac{4}{5} = 1\). Hence, \(P(A \cup B)\) can be equal to or less than 1, but not necessarily \(\geq \frac{4}{5}\). Hence, option (A) might not always hold.

Step 3: Analyze \(P(A \cap B)\)

Since the maximum value for \(P(A \cap B)\) can be \(P(A)\), we have \(P(A \cap B) \leq \frac{2}{5}\). The minimum value could be \(0\), though typically, since \(A\) and \(B\) both exist, \(\frac{1}{5} \leq P(A \cap B)\) can hold if the events are not mutually exclusive. Thus, option (B) holds true.

Step 4: Confirm \( \frac{1}{4} \leq P(A / B) \leq \frac{1}{2} \)

By definition, \(P(A / B) = \frac{P(A \cap B)}{P(B)}\). We previously bounded \(\frac{1}{5} \leq P(A \cap B) \leq \frac{2}{5}\). Thus, \[\frac{1}{5} \times \frac{5}{4} \leq P(A / B) = \frac{P(A \cap B)}{\frac{4}{5}} \leq \frac{2}{5} \times \frac{5}{4}\], leading to \(\frac{1}{4} \leq P(A / B) \leq \frac{1}{2}\). Option (C) is correct.

Step 5: Evaluate \(P\left(A \cap B^{\prime}\right)\)

Since \(P(A \cap B)\) can be maximized to \(\frac{2}{5}\), \[P(A \cap B^{\prime}) = P(A) - P(A \cap B) \geq \frac{2}{5} - \frac{2}{5} = 0\]. Therefore, \(P(A \cap B^{\prime})\) can be zero or more but not necessarily \(\leq \frac{1}{5}\). Option (D) might not hold.

Key Concepts

Union of Events

The union of events is a fundamental concept in probability that helps us understand the likelihood of either one event happening or both events happening together. This is denoted as \( A \cup B \), which means "either A happens, B happens, or both happen." Essentially, it represents the combination of all possible outcomes from both events without double counting any overlaps.
To calculate the probability of the union of two events \( A \) and \( B \), we use the formula:

• \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)

This formula accounts for the probability of both events happening at the same time (the intersection) to avoid counting it twice.
The union allows us to assess broader possibilities, crucial in scenarios where multiple outcomes are acceptable or desired.

Intersection of Events

The intersection of events zooms into the specific outcome where two events happen simultaneously. If you think about events \( A \) and \( B \) happening together, the intersection, denoted as \( A \cap B \), gives the probability that both events occur at the same time.
The formula for the intersection is straightforward:

• \( P(A \cap B) \)

This focuses on the overlap between \( A \) and \( B \).
In our exercise, we established a range for \( P(A \cap B) \), showing that this probability lies between \( \frac{1}{5} \) and \( \frac{2}{5} \) when the events are not mutually exclusive.
Understanding intersections is key in problems where simultaneous outcomes are of interest, such as analyzing overlapping customer demographics or concurrent events.

Conditional Probability

Conditional probability deals with the likelihood of an event occurring given that another event has already occurred. This relationship is shown as \( P(A / B) \), which reads "the probability of A given B." It represents scenarios where the occurrence of one event impacts the probability of another.
The calculation of conditional probability involves this formula:

• \( P(A / B) = \frac{P(A \cap B)}{P(B)} \)

This equation tells us how our initial opinion of an event (\( A \)) changes when we know that \( B \) has occurred.
In the exercise, this was used to confirm that the conditional probability \( \frac{1}{4} \leq P(A / B) \leq \frac{1}{2} \) was correct, given the identified range for \( P(A \cap B) \).
Grasping conditional probability concepts is critical in areas like risk assessment and decision making, where prior information can influence outcomes.

Complement of an Event

The complement of an event encapsulates the idea of an event not happening. If event \( A \) is something we can quantify, then \( A' \) (pronounced "A prime") is the event of \( A \) not occurring. This is a part of the probability basics where we explore alternatives to the happening event.
For complementary events, the rule is simple but essential:

• \( P(A') = 1 - P(A) \)

This rule complements our understanding by filling in the gap to total probability, ensuring everything sums up to one.
In our scenario, we analyzed \( P(A \cap B') \), which considered the probability of \( A \) happening without \( B \) happening. This demonstrated the logical thinking used to dissect various outcome combinations.
Understanding complements is essential when analyzing non-occurrence and its implications, offering a complete view of likelihoods in probability theory.