Problem 125 - 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

(A) \( p = 1 \) and (D) \( q = 0 \) are correct conditions.

Step 1: Understanding the Problem

To solve, we need to understand the conditions under which the student is successful: he passes test I & II, or test I & III. We use the given probabilities to determine under these conditions what the values of \( p \) and \( q \) must be to meet the total success probability of \( \frac{1}{2} \).

Step 2: Expressing the Problem Mathematically

Define the probabilities: \( P(\text{I and II}) = pq \), \( P(\text{I and III}) = p \times \frac{1}{2} \). The student is successful if he passes either I and II, or I and III. Since these are mutually exclusive, the probability can be written as \( P(S) = pq + p \cdot \frac{1}{2} = \frac{1}{2} \).

Step 3: Substitution and Simplification

Substitute \( P(S) = \frac{1}{2} \) into the equation for student success:\[ pq + \frac{p}{2} = \frac{1}{2} \]Simplify to find a relation between \( p \) and \( q \). Multiply through by 2 to eliminate the fraction:\[ 2pq + p = 1 \]

Step 4: Analyzing Possible Solutions

Consider each given option.1. \( p = 1 \): Substitute into \( 2pq + p = 1 \) - \( 2q + 1 = 1 \) gives \( q = 0 \).2. \( q = 1 \): Substitute into \( 2pq + p = 1 \) - \( 2p + p = 1 \) gives \( 3p = 1 \) which is not possible since \( p \leq 1 \).Thus, only \( p = 1 \) and \( q = 0 \) satisfies the requirement.

Key Concepts

Conditional Probability

In probability theory, conditional probability is all about finding the chance that an event will occur, given that we already have some information about another related event. To understand this concept, imagine taking a test. Suppose you know that passing Test I makes a student 70% likely to pass Test II. The probability of passing Test II now depends on knowing that Test I has been passed. That's the essence of conditional probability.
This is often denoted with \( P(A|B) \), which reads as "the probability of A given B." This tells us how likely A is after knowing B has occurred. In our exercise, the student’s success probability was determined by passing different combinations of tests, given the underlying probabilities of passing each test individually.
The conditional element comes from knowing which tests are passed, which affects overall success. This idea helps in understanding how events influence each other, allowing us to predict outcomes based on specific conditions.

Mutually Exclusive Events

Mutually exclusive events are scenarios where two or more events cannot occur at the same time. Imagine flipping a coin: you can either get heads or tails, but not both. In the context of our problem, we can think of it as passing Test I and II or passing Test I and III.
These are mutually exclusive because passing both tests simultaneously is not considered. Therefore, if the student passes one set of tests, they haven’t passed the other set at that moment, simplifying the total probability calculation.
When dealing with mutually exclusive events, the probability of either event happening is simply the sum of their individual probabilities. So, the formula is \( P(A \, \text{or} \, B) = P(A) + P(B) \).
In problems such as these, understanding whether events are mutually exclusive helps to directly apply this additivity in probabilities, which makes calculations easier and faster.

Success Probability

Success probability refers to the likelihood of achieving a desired outcome. In many cases, this means computing how likely it is for a combination of conditions to result in a favorable scenario. In this exercise, the student is deemed successful if certain combinations of test results occur. Understanding success probability helps to focus on what is truly needed to reach the desired outcome.
Let's say success is defined by passing Tests I and II, or Tests I and III. Given the probability of passing each test, success probability combines them based on these conditions.
This is calculated by determining individual event probabilities and how they combine, as seen with the equation \( pq + \frac{p}{2} \) being set to \( \frac{1}{2} \). By doing so, you logically determine which probabilities meet the target success criterion. Success probability also encourages exploring alternate pathways of achieving the result, given different conditions.