Problem 124 - A Complete Resource Book in Mathematics for JEE Main 2017

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

(D) \(q=0\)

Step 1: Identify the Events

Define the events for passing each test. Let event A be passing test I, event B be passing test II, and event C be passing test III. The probabilities are given as \(P(A) = p\), \(P(B) = q\), and \(P(C) = \frac{1}{2}\).

Step 2: Define Success Criteria

The student is considered successful if they pass either test I and II together, or test I and III together. Thus, success can be described by the union of two intersections: (i) passing test I and II, which is \(A \cap B\) and (ii) passing test I and III, which is \(A \cap C\). The event of being successful is \((A \cap B) \cup (A \cap C)\).

Step 3: Apply Probability Theorem

Using the addition law of probability for the union of events, the probability of being successful is \(P((A \cap B) \cup (A \cap C)) = P(A \cap B) + P(A \cap C) - P(A \cap B \cap C)\).

Step 4: Substitute Known Values

Substitute the known values for the individual probabilities: \(P(A \cap B) = p \cdot q\), \(P(A \cap C) = p \cdot \frac{1}{2}\), and \(P(A \cap B \cap C) = p \cdot q \cdot \frac{1}{2}\).

Step 5: Set Up the Probability Equation

Equate the expression for successful probability to \(\frac{1}{2}\): \[p \cdot q + p \cdot \frac{1}{2} - p \cdot q \cdot \frac{1}{2} = \frac{1}{2}.\]

Step 6: Simplify the Equation

Combine like terms and simplify: \[p \cdot q + \frac{p}{2} - \frac{p \cdot q}{2} = \frac{1}{2}.\] Multiply through by 2 to clear fractions: \[2pq + p - pq = 1.\]

Step 7: Further Simplification

Simplify the expression further by recognizing a common factor: \[pq + p = 1.\] Factor out \(p\): \[p(q + 1) = 1.\]

Step 8: Solve for p or q

Setting the factored expression equal gives us \(p(q + 1) = 1\). Since \(q + 1\) must be non-zero, rearrange to find \(p = \frac{1}{q+1}\). Since \(p\) must be either 0 or 1 under given options, \(q\) must make this equality meaningful.If \(q = 0\), then \(p(1) = 1\) implies \(p = 1\). Test other options to validate.

Key Concepts

Addition Law of Probability

The addition law of probability is a fundamental rule used to find the probability of the union of two or more events. It helps us understand how to combine probabilities of different events occurring. Mathematically, for two events, A and B, the addition law states that the probability of either A or B occurring is given by:

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

This formula accounts for both events occurring at once. If you add the probabilities of A and B separately, you count the intersection of A and B twice, once in \(P(A)\) and once in \(P(B)\). Therefore, you need to subtract \(P(A \cap B)\) to get an accurate probability.

In our exercise, we deal with three events (tests) and use the extended version of the addition law to calculate the probability of success based on combinations of these tests.

Union of Events

The union of events means that we consider the occurrence of at least one of several events. In probability terms, the union of two events, A and B, denoted \(A \cup B\), represents the scenario where either event A, event B, or both happen.

Let's apply this to the student's scenario: the student is considered successful if they pass either test I and II or test I and III. Therefore, the event of being successful is presented as a union of two intersections:

• Passing test I and II: \(A \cap B\)
• Passing test I and III: \(A \cap C\)

The union of these events, \((A \cap B) \cup (A \cap C)\), covers both conditions for success, allowing us to use the addition law of probability to find the total probability of the student being successful.

Intersection of Events

The intersection of events, indicated as \(A \cap B\), represents the situation where all involved events occur simultaneously. This is akin to looking for an overlap between events. For example, in our scenario, the intersection \(A \cap B\) means the student passes both test I and test II.

We use this concept primarily to determine precisely when multiple criteria are met together. In our problem, two key intersections define success:

• Passing test I and II: \(A \cap B\)
• Passing test I and III: \(A \cap C\)

Additionally, within the formulation, we also consider the intersection of all three events, \(A \cap B \cap C\), which would occur if the student passes all tests I, II, and III. Understanding intersections helps us accurately calculate the likelihood of combined event outcomes crucial in multi-event probability scenarios.