Problem 68 - A Complete Resource Book in Mathematics for JEE Main

Question

It is given that the events \(A\) and \(B\) are such that \(P(A)=\) \(\frac{1}{4}, P\left(\frac{A}{B}\right)=\frac{1}{2}\) and \(P\left(\frac{B}{A}\right)=\frac{2}{3} .\) Then \(P(B)\) is (A) \(1 / 6\) (B) \(1 / 3\) (C) \(2 / 3\) (D) \(1 / 2\)

Step-by-Step Solution

Answer

The probability \( P(B) \) is \( \frac{1}{3} \).

Step 1: Extract Given Information

The problem provides the following probabilities: \( P(A) = \frac{1}{4} \), \( P(A \mid B) = \frac{1}{2} \), and \( P(B \mid A) = \frac{2}{3} \). We are asked to find \( P(B) \).

Step 2: Relate Conditional Probabilities with Joint Probability

Utilize the definition of conditional probability to write expressions for the given conditional probabilities: \( P(A \mid B) = \frac{P(A \cap B)}{P(B)} \) and \( P(B \mid A) = \frac{P(A \cap B)}{P(A)} \).

Step 3: Calculate Probability \( P(A \cap B) \)

Using the equation \( P(B \mid A) = \frac{P(A \cap B)}{P(A)} \), substitute the known values: \( \frac{2}{3} = \frac{P(A \cap B)}{\frac{1}{4}} \). Solve for \( P(A \cap B) \) to find it is \( \frac{2}{3} \times \frac{1}{4} = \frac{1}{6} \).

Step 4: Solve for \( P(B) \)

Use the expression for conditional probability \( P(A \mid B) = \frac{P(A \cap B)}{P(B)} \) again. Substitute \( \frac{1}{2} = \frac{\frac{1}{6}}{P(B)} \) and solve for \( P(B) \). This gives \( P(B) = \frac{1}{6} \div \frac{1}{2} = \frac{1}{3} \).

Step 5: Identify Correct Answer

The calculated value of \( P(B) = \frac{1}{3} \). Based on the multiple-choice options, the correct answer is (B) \( \frac{1}{3} \).

Key Concepts

Conditional Probability

Conditional probability is a key concept in probability theory that expresses the likelihood of an event occurring given that another event has already occurred. Imagine it's raining, and you want to know the chance of seeing a rainbow. The probability of seeing a rainbow given that it's raining is an example of conditional probability. It's expressed as \( P(A \mid B) \), where \( A \) and \( B \) are events, and it's calculated by the formula \( P(A \mid B) = \frac{P(A \cap B)}{P(B)} \).
In the given exercise, the conditional probability \( P(A \mid B) = \frac{1}{2} \) means that, if event \( B \) has occurred, the probability that \( A \) will occur is \( 50\% \). Understanding conditional probability helps in breaking down complex probability scenarios into more manageable pieces.

Joint Probability

Joint probability refers to the probability of two events occurring together, symbolized as \( P(A \cap B) \). This concept is crucial when you need to determine the probability of two or more events happening at the same time.
For instance, consider tossing two coins and finding the probability of both landing heads. This scenario involves joint probability, denoted as \( P(A \cap B) = P(A) \times P(B) \) if the events are independent.
In the exercise, we have already calculated the joint probability \( P(A \cap B) = \frac{1}{6} \) using the conditional probability formula. Practically, this suggests that the chance of both events \( A \) and \( B \) occurring is \( \frac{1}{6} \), or about \( 16.67\% \). Joint probability enables us to understand how events interplay and affect their combined outcome.

Concept of Events

Understanding events in probability is foundational to grasping more complex concepts like joint and conditional probabilities. An 'event' in probability is essentially a set of outcomes from an experiment.
For example, consider rolling a die. Rolling an even number, such as \( \{2, 4, 6\} \), is an event. Events can be independent, dependent, mutually exclusive, or inclusive, affecting how their probabilities are calculated.
In our exercise, we define events \( A \) and \( B \), where knowing some probabilities, such as \( P(A) = \frac{1}{4} \), helps build the framework to solve for unknowns like \( P(B) \).
Recognizing the type and nature of events is crucial for correctly applying probability rules, such as conditional probability, ensuring accurate calculations and deeper comprehension of probability scenarios.