Problem 124 - A Complete Resource Book in Mathematics for JEE Main

Question

A student appears for test I, II and III. The student is successful if he passes either in test I and II or test I and III. The probability of the student passing in tests I, II and III are \(p, q\) and \(\frac{1}{2}\), respectively. If the probability that the student is successful is \(\frac{1}{2}\), then (A) \(p=1\) (B) \(p=0\) (C) \(q=1\) (D) \(q=0\)

Step-by-Step Solution

Answer

The correct answer is (D) \(q=0\).

Step 1: Identify the condition for success

The student is successful if he passes either in tests I and II or in tests I and III. Define this event as the union of two events: event A (passing tests I and II) and event B (passing tests I and III).

Step 2: Define the probabilities of the events

Event A corresponds to passing tests I and II, which has a probability of passing denoted by \(pq\). Event B corresponds to passing tests I and III, which has a probability denoted by \(p \cdot \frac{1}{2}\).

Step 3: Express the overall success probability

The probability of success is the union of events A and B. Using the principle of inclusion-exclusion for two events, the probability is \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).

Step 4: Calculate the intersection probability

Since passing test I is common in both events A and B, the intersection probability is \(P(A \cap B) = p \cdot q \cdot \frac{1}{2}\).

Step 5: Apply the given success probability

We are given that the probability of success is \(\frac{1}{2}\). Setting the events equation to this probability, we have \(pq + \frac{p}{2} - \frac{pq}{2} = \frac{1}{2}\).

Step 6: Simplify and solve the equation

Multiply through by 2 to eliminate the fraction: \(2pq + p - pq = 1\). Simplify the expression to \(p(2q + 1) - pq = 1\) or \(pq + p = 1\).

Step 7: Solve for p in terms of q

From the equation \(p(q + 1) = 1\), we can deduce that \(p = \frac{1}{q + 1}\).

Step 8: Determine possible values for q

For \(p\) to be a valid probability (i.e., between 0 and 1), \(\frac{1}{q+1}\) must be within that range. The options provided suggest integer values for \(q\).

Step 9: Analyze given options

Substitute the options into \(p = \frac{1}{q+1}\):- If \(q=1\), then \(p = \frac{1}{2}\) (which is valid).- If \(q=0\), then \(p = 1\) (which is also valid).

Step 10: Conclusion based on analysis

Compare with the options: If \(q=0\), then it fits the condition of a maximal valid \(p\). Therefore, the option that ensures the condition is \(q=0\). This supports option (D).

Key Concepts

Inclusion-Exclusion Principle

The Inclusion-Exclusion Principle is an essential rule in probability theory that helps in accurately determining the probability of the union of multiple events. It accounts for the probabilities of individual events, while also correcting for any overlapping probabilities, ensuring no probabilities are counted more than once. In this exercise, we use the principle to find the probability that the student either passes tests I and II or tests I and III. More formally, if we have events A and B, the probability that either A or B occurs is given by:\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]This means we add together the probabilities of A and B occurring separately but then subtract the probability of them occurring together, which accounts for double-counting those scenarios where both occur at the same time. So, when applying this technique to our problem:

• The probability of passing both Test I and Test II is \(P(A) = pq\).
• The probability of passing both Test I and Test III is \(P(B) = p \cdot \frac{1}{2}\).
• The probability of passing all three tests (intersection) is \(P(A \cap B) = p \cdot q \cdot \frac{1}{2}\).

Understanding and applying this principle correctly helps solve complex probability problems by breaking them into simpler events.

Events in Probability

In probability theory, events represent outcomes or results from a particular situation or experiment. For example, in this exercise, the events are defined based on passing certain combinations of tests. It is crucial to clearly define these events to calculate probabilities effectively. In this context:

• Event A is defined as passing Test I and Test II. Mathematically, if Test I's probability is \(p\) and Test II's probability is \(q\), then the probability of Event A is \(P(A) = pq\).
• Event B is defined as passing Test I and Test III. Here, given that the probability of passing Test III is \(\frac{1}{2}\), the probability of Event B is \(P(B) = p \cdot \frac{1}{2}\).

Defining events this way makes it easier to understand and solve the entire problem, as each event represents a clear and identifiable outcome related to the student's performance in these tests. This enables us to use structured formulas and logical steps to find the solution.

Conditional Probability

Conditional probability is a way to measure the likelihood of an event occurring, given that another event has already happened. In simple terms, it helps us understand how the probability of an event changes when we know some other information beforehand. In this exercise, while not explicitly stated, conditional probability concepts indirectly play a role. The student passing tests I and II or I and III ties the success conditions to certain dependencies between the tests. This intertwines with the rules of conditional probabilities, as the success in one test affects the understanding of the probability of passing another test. Consider an example: knowing the student has passed Test I changes how we assess their chance of success in tests II or III. Although the problem itself doesn't directly ask us to calculate a conditional probability, understanding these indirect relationships helps in interpreting how different probabilities interact and add up according to the conditions given in the problem. Handling dependent events correctly and understanding these dependencies is a key component in solving the problem accurately. Conditional probability principles guide us in simplifying such interdependent scenarios.