Problem 26 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

For two events \(A\) and \(B\), if \(P(A)=P\left(\frac{A}{B}\right)=\frac{1}{4}\) and \(P\left(\frac{B}{A}\right)=\frac{1}{2}\), then (a) \(A\) and \(B\) are independent (b) \(P\left(\frac{A^{\prime}}{B}\right)=\frac{3}{4}\) (c) \(P\left(\frac{B^{\prime}}{A^{\prime}}\right)=\frac{1}{2}\) (d) All of these

Step-by-Step Solution

Answer

The correct answer is (d) All of these.

Step 1: Understand the Given Probabilities

We are given two events, \( A \) and \( B \), with the probabilities \( P(A)=\frac{1}{4} \), \( P\left(\frac{A}{B}\right)=\frac{1}{4} \), and \( P\left(\frac{B}{A}\right)=\frac{1}{2} \). "\( P\left(\frac{A}{B}\right) \)" refers to the conditional probability of \( A \) given \( B \), and similarly, \( P\left(\frac{B}{A}\right) \) refers to the conditional probability of \( B \) given \( A \).

Step 2: Use Conditional Probability Formulas

The conditional probability \( P\left(\frac{A}{B}\right) = \frac{P(A \cap B)}{P(B)} \) and \( P\left(\frac{B}{A}\right) = \frac{P(A \cap B)}{P(A)} \). From the information given, \( \frac{P(A \cap B)}{P(B)} = \frac{1}{4} \) and \( \frac{P(A \cap B)}{P(A)} = \frac{1}{2} \).

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

From \( P\left(\frac{B}{A}\right)=\frac{1}{2} \), we have \( \frac{P(A \cap B)}{P(A)} = \frac{1}{2} \). Since \( P(A)=\frac{1}{4} \), \( P(A \cap B) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8} \).

Step 4: Calculate \( P(B) \)

As \( P\left(\frac{A}{B}\right)=\frac{1}{4} \), this gives \( \frac{P(A \cap B)}{P(B)} = \frac{1}{4} \). Given \( P(A \cap B)=\frac{1}{8} \), then \( P(B) = \frac{1}{8} \times 4 = \frac{1}{2} \).

Step 5: Check Independence of \( A \) and \( B \)

Two events are independent if \( P(A \cap B) = P(A)P(B) \). Here, \( P(A \cap B) = \frac{1}{8} \) and \( P(A)P(B) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} \). Since these are equal, \( A \) and \( B \) are independent.

Step 6: Calculate \( P\left(\frac{A^{\prime}}{B}\right) \)

\( P\left(\frac{A^{\prime}}{B}\right) = 1 - P\left(\frac{A}{B}\right) = 1 - \frac{1}{4} = \frac{3}{4} \).

Step 7: Calculate \( P\left(\frac{B^{\prime}}{A^{\prime}}\right) \)

We know \( P(A\cup B) = P(A) + P(B) - P(A\cap B) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8} = \frac{5}{8} \). So, \( P(A^{\prime}\cap B^{\prime}) = 1 - P(A\cup B) = 1 - \frac{5}{8} = \frac{3}{8} \). This implies \( P(A^{\prime})\) is \( 1 - P(A)=\frac{3}{4}\). Then, \( P\left(\frac{B^{\prime}}{A^{\prime}}\right) = \frac{P(A^{\prime}\cap B^{\prime})}{P(A^{\prime})} = \frac{3/8}{3/4} = \frac{1}{2} \).

Step 8: Conclusion

Based on the above calculations, all the statements (a), (b), and (c) are true.

Key Concepts

Probability of Complementary Events

The concept of complementary events in probability is quite intriguing. A complementary event refers to the occurrence of the event that does not happen. If we have an event "A", the complement of this event, denoted as "A'", includes all outcomes in the sample space that are not in "A". The probability of a complementary event can be found using the formula:

• \( P(A') = 1 - P(A) \)

This indicates that the sum of the probabilities of an event and its complement is always one. For instance, if the probability of raining tomorrow is 0.3, the probability it doesn’t rain (the complementary event) would be 0.7.

In our exercise, we have utilized this idea to find the probability of the complement of event A given B:

• \( P\left( \frac{A'}{B} \right) = 1 - P\left( \frac{A}{B} \right) \)
• Evaluating, we have \( P\left( \frac{A'}{B} \right) = \frac{3}{4} \).

Understanding complementary probabilities can vastly simplify many problems, especially when working with more complex events.

Independence of Events

Determining whether two events are independent is essential in probability theory. Two events, A and B, are independent if the occurrence of one does not affect the occurrence of the other. Formally, A and B are independent if:

• \( P(A \cap B) = P(A) \times P(B) \)

This equality means that the joint probability of both events happening is simply the product of their individual probabilities.

In the exercise, after calculating both separate and joint probabilities, we found:

• \( P(A) = \frac{1}{4} \), \( P(B) = \frac{1}{2} \), \( P(A \cap B) = \frac{1}{8} \)
• \( P(A) \times P(B) = \frac{1}{8} \)

Since these two values match, it confirms that events A and B are indeed independent.

This independence allows us to make predictions or calculate other probabilities without considering the effects of one event on the other.

Conditional Probability Formula

Conditional probability helps us update the likelihood of an event based on the occurrence of another event. The conditional probability of A given B helps us understand the probability of A occurring once we know B has happened. The formula is:

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

This is a crucial tool in probability as it allows for the division of joint probability with the probability of the condition.

In the given exercise, we calculated the conditional probability of A given B as \( \frac{1}{4} \) and conversely B given A as \( \frac{1}{2} \). These illustrate how knowledge of one event affects the consideration of another.

Next, the ability to transition between joint & conditional probabilities is foundational in probability theory, as it is used across various statistical models and real-life scenarios where conditions play a crucial role.