Problem 8 - A Complete Resource Book in Mathematics for JEE Main

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) \(A\) is subevent of \(B\) (B) \(A\) and \(B\) are mutually exclusive (C) \(A\) and \(B\) are independent and \(P\left(\frac{A^{\prime}}{B}\right)=\frac{3}{4}\) (D) none of the above

Step-by-Step Solution

Answer

Option (C): \(A\) and \(B\) are independent and \(P\left(\frac{A'}{B}\right) = \frac{3}{4}\).

Step 1: Understand Given Probabilities

We are given that the probability of event \(A\) is \(P(A) = \frac{1}{4}\). We are also given the conditional probabilities \(P\left(\frac{A}{B}\right) = \frac{1}{4}\) and \(P\left(\frac{B}{A}\right) = \frac{1}{2}\). These probabilities imply particular relationships between events \(A\) and \(B\).

Step 2: Express Conditionals Using Basic Probability Rules

From probability rules, \(P\left(\frac{A}{B}\right) = \frac{P(A \cap B)}{P(B)} = \frac{1}{4}\). Similarly, \(P\left(\frac{B}{A}\right) = \frac{P(A \cap B)}{P(A)} = \frac{1}{2}\).

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

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

Step 4: Compare Joint Probability with \(P\left(\frac{A}{B}\right)\) Equation

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

Step 5: Analyze Independence

For \(A\) and \(B\) to be independent, \(P(A \cap B) = P(A) \times P(B)\) must hold. Here, \(\frac{1}{8} = \frac{1}{4} \times \frac{1}{2}\) which is true, so \(A\) and \(B\) are independent.

Step 6: Verify Complementary Conditional Probability

If \(A\) and \(B\) are independent, then \(P\left(\frac{A^{\prime}}{B}\right) = 1 - P\left(\frac{A}{B}\right) = 1 - \frac{1}{4} = \frac{3}{4}\). This is consistent with option (C).

Key Concepts

Independence of Events

When two events are independent, the occurrence of one event does not affect the probability of the other event occurring. For mathematical clarity, events \(A\) and \(B\) are independent if the probability of both events happening together, known as the joint probability \(P(A \cap B)\), equals the product of their individual probabilities:

• \(P(A \cap B) = P(A) \times P(B)\)

In our exercise, we have \(P(A) = \frac{1}{4}\) and \(P(B) = \frac{1}{2}\). We calculated \(P(A \cap B)\) to be \(\frac{1}{8}\), which matches the product of \(\frac{1}{4}\) and \(\frac{1}{2}\). Thus, events \(A\) and \(B\) are indeed independent, since their joint probability calculates correctly with their independence condition.

Joint Probability

Joint probability refers to the probability of two events happening at the same time. For events \(A\) and \(B\), it is represented as \(P(A \cap B)\).

• It is vital for understanding how events relate to one another when they are not independent.
• Joint probability is calculated by rearranging conditional probabilities or by finding the overlap of both events on a probability scale.

In this scenario, we calculated \(P(A \cap B)\) in two ways: - From the conditional probability \(P\left(\frac{B}{A}\right) = \frac{1}{2}\), which led us to \(P(A \cap B) = \frac{1}{8}\).- Also through \(P\left(\frac{A}{B}\right) = \frac{1}{4}\) and solving for \(P(B)\), reinforcing our understanding of the event overlaps.

Mutually Exclusive Events

Mutually exclusive events are events that cannot happen at the same time. If \(A\) and \(B\) are mutually exclusive, then \(P(A \cap B) = 0\).

• For instance, flipping a coin can result in either heads or tails, but not both simultaneously.
• Mutually exclusive implies if one event happens, the other cannot.

In the exercise at hand, we established that \(P(A \cap B) = \frac{1}{8}\), proving that \(A\) and \(B\) are not mutually exclusive. They can occur together, as evidenced by their calculated joint probability, distinguishing them from exclusive events.