Problem 131 - A Complete Resource Book in Mathematics for JEE Main 2017

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

All options A, B, C, and D are correct based on the probability calculations.

Step 1: Clarify the Problem Statement

We need to evaluate the given alternatives (A, B, C, D) using the probability rules. The probabilities provided are \(P(A) = \frac{2}{5}\) and \(P(B) = \frac{4}{5}\). We will use these to determine the measure of each option given.

Step 2: Evaluate Option A

To check if \(P(A \cup B) \geq \frac{4}{5}\), use the formula: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Since \(P(A \cap B)\) is unknown, let's set its possible range: \(0 \leq P(A \cap B) \leq \min(P(A), P(B)) = \frac{2}{5}\). Find the maximum possible value when \(P(A \cap B) = 0\): \[ P(A \cup B) = \frac{2}{5} + \frac{4}{5} - 0 = \frac{6}{5} \] This satisfies \(P(A \cup B) \geq \frac{4}{5}\).

Step 3: Evaluate Option B

To verify:\( \frac{1}{5} \leq P(A \cap B) \leq \frac{2}{5} \), use the intersection probability range: Since \(P(A) + P(B) = \frac{6}{5}\), and using \(P(A \cup B) \leq 1\), we find: \[ P(A \cap B) = P(A) + P(B) - P(A \cup B) \geq \frac{2}{5} + \frac{4}{5} - 1 = \frac{1}{5} \] Thus, \(\frac{1}{5} \leq P(A \cap B) \leq \frac{2}{5}\) is valid.

Step 4: Evaluate Option C

Use this formula: \[ P(A / B) = \frac{P(A \cap B)}{P(B)} \]Assume \(P(A \cap B)\) at its range lower end, \(\frac{1}{5}\), and upper end, \(\frac{2}{5}\): - Lower end: \(\frac{\frac{1}{5}}{\frac{4}{5}} = \frac{1}{4}\) - Upper end: \(\frac{\frac{2}{5}}{\frac{4}{5}} = \frac{1}{2}\)Thus, \(\frac{1}{4} \leq P(A / B) \leq \frac{1}{2}\) is accurate.

Step 5: Evaluate Option D

Check if \(P(A \cap B') \leq \frac{1}{5}\):\[ P(A \cap B') = P(A) - P(A \cap B) \], thus - Maximum \(P(A \cap B) = \frac{2}{5}\), then \(\frac{2}{5} - \frac{2}{5}= 0\),- Minimum \(P(A \cap B) = \frac{1}{5}\), then \(\frac{2}{5} - \frac{1}{5}= \frac{1}{5}\)Therefore, \(P(A \cap B') \leq \frac{1}{5}\) is valid.

Key Concepts

Conditional Probability

Conditional probability helps us understand how the probability of an event changes when another event is known to have occurred. It's a way of updating the likelihood of an event based on new information. Mathematically, the conditional probability of an event \(A\) given an event \(B\) is denoted as \(P(A / B)\) and is calculated using the formula:

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

This formula tells us that to find the conditional probability, we divide the probability of the intersection of both events by the probability of the event that we are conditioning on. For instance, in the exercise, we calculated \(P(A / B)\) to ensure it lies between \(\frac{1}{4}\) and \(\frac{1}{2}\), meaning that the probability of \(A\) occurring when \(B\) is certain falls within those bounds. This illustrates the concept that probabilities can be dependent on conditions.

Union of Events

In probability, the union of events refers to the set containing all outcomes that belong to either one of the events or to both. If you imagine sets, the union is where you combine all elements from each set without duplicating any elements. For two events \(A\) and \(B\), the probability of their union is represented as \(P(A \cup B)\). The formula to calculate this is:

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

This formula reflects the idea of adding the probabilities of each event and then subtracting the probability of their intersection because it is counted twice. In the context of the exercise, we interpreted \(P(A \cup B)\) and found that it should be greater than or equal to \(\frac{4}{5}\). Our evaluation showed that, given the certain probabilities of \(A\) and \(B\), this criterion is satisfied.

Intersection of Events

The intersection of events is a fundamental idea where we look at the probability of two events happening at the same time. For example, the intersection of \(A\) and \(B\) is represented as \(A \cap B\), which focuses on the outcomes common to both events. In our problem, we figured out that the probability of the intersection falls between \(\frac{1}{5}\) and \(\frac{2}{5}\). This was done using the formula for union of events and ensuring the total probability doesn’t exceed 1. Thus, the intersection tells us how much overlap there is between the two events. This range of values (between \(\frac{1}{5}\) and \(\frac{2}{5}\)) provides insight into how often both \(A\) and \(B\) happen together, within the limits given in the exercise.

Complementary Events

Complementary events are all about understanding what happens outside of a specific event. If \(A\) is an event, its complement \(A^\prime\) consists of all outcomes not in \(A\). The probability of \(A^\prime\) is obtained by subtracting \(P(A)\) from 1:

• \(P(A^\prime) = 1 - P(A)\)

In our exercise, we explored complementary events by examining \(P(A \cap B^\prime)\), which refers to \(A\) occurring without \(B\) occurring. The calculation showed us that \(P(A \cap B^\prime)\) is indeed less than or equal to \(\frac{1}{5}\). Complementary events highlight what's not included within an event, helping provide a full picture of the probabilities involved.