Problem 13 - 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) \(p=1, q=\frac{1}{2}\) (B) \(p=1, q=0\) (C) \(p=q=1 \frac{1}{2}\) (D) \(p=q=1\).

Step-by-Step Solution

Answer

(B) \( p=1, q=0 \).

Step 1: Identifying the Probabilities

Identify the events for the student to be successful. The student passes the tests if they pass Test I & II, or Test I & III. Let's denote these events as follows: - Let event A be passing Test I, - Event B be passing Test II,- Event C be passing Test III. We are given: - Probability of passing Test I, P(A) = p,- Probability of passing Test II, P(B) = q,- Probability of passing Test III, P(C) = \( \frac{1}{2} \).

Step 2: Writing the Success Condition

The student is successful if they pass Test I & II, or Test I & III. This can be written as the union of events (A \( \cap \) B) or (A \( \cap \) C). So,\[ \text{Successful} = (A \cap B) \cup (A \cap C) \]According to the probability union rule:\[ P((A \cap B) \cup (A \cap C)) = P(A \cap B) + P(A \cap C) - P((A \cap B) \cap (A \cap C)) \]

Step 3: Calculating Individual Probabilities

Now calculate each component of this equation:- P(A \( \cap \) B) = P(A) \( \times \) P(B) = p \( \times \) q.- P(A \( \cap \) C) = P(A) \( \times \) P(C) = p \( \times \) \( \frac{1}{2} \) = \( \frac{p}{2} \).- P((A \( \cap \) B) \( \cap \) (A \( \cap \) C)) = P(A \( \cap \) B \( \cap \) C) = P(A) \( \times \) P(B) \( \times \) P(C) = p \( \times \) q \( \times \) \( \frac{1}{2} \) = \( \frac{pq}{2} \).

Step 4: Calculating the Probability of Being Successful

Substituting these in the union rule:\[ P((A \cap B) \cup (A \cap C)) = pq + \frac{p}{2} - \frac{pq}{2} \]We also know that the probability of the student being successful is \( \frac{1}{2} \). So,\[ pq + \frac{p}{2} - \frac{pq}{2} = \frac{1}{2} \]

Step 5: Solving the Equation

Combine and simplify the equation:\[ pq - \frac{pq}{2} + \frac{p}{2} = \frac{1}{2} \]Multiply each term by 2 to eliminate the fraction:\[ 2pq - pq + p = 1 \]This simplifies to:\[ pq + p = 1 \]Factor out common term:\[ p(q + 1) = 1 \]From this, we get:\[ p = 1 \] and hence:\[ q + 1 = 1 \rightarrow q = 0 \]

Step 6: Choosing the Correct Option

According to the given options, the values that satisfy the conditions are \( p = 1 \) and \( q = 0 \), which corresponds to:- Option (B): \( p=1, q=0 \).

Key Concepts

Conditional Probability

Conditional probability is a fascinating concept that helps us understand the likelihood of an event occurring, given that another event has already occurred. This type of probability is denoted as \( P(A|B) \), and reads as "the probability of event A given event B." In the context of our original exercise, while conditional probability is not explicitly calculated, understanding it deepens our comprehension of the sequence of dependent events.

Let's break down how conditional probability operates. When calculating \( P(A|B) \), you're considering only those outcomes where B happens and asking how likely A is, within that subset. The formula can be expressed as:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
This formula shows how conditional probability narrows our focus to a more specific scenario by dividing the probability of both events occurring by the likelihood of the given condition, B.

In real-world terms, consider a simple example: If you're choosing cards from a deck, and someone tells you the card is a heart, the conditional probability would calculate the chance of next drawing a king given that it's a heart, based only on the remaining hearts in the deck. This principle is incredibly useful in scenarios where events are dependent on each other.

Event Intersection

Event intersection, denoted \( A \cap B \), is a concept that shows the overlapping occurrence of two or more events. In probability, it helps in identifying situations where multiple conditions are simultaneously true. For example, in our exercise, the student needs to pass both Test I & II, represented as \( A \cap B \), to be successful.

Calculating the probability where two events intersect involves multiplying their individual probabilities if they are independent. The formula is:
\[ P(A \cap B) = P(A) \times P(B) \]
However, if they are dependent events, the calculation might involve conditional probabilities, as discussed earlier.

It's essential to understand event intersections when solving probability problems because it represents all possible outcomes satisfying both scenarios. Knowing this concept helps predict outcomes in many real-life applications, like determining how likely it is for both the sun to shine and a bus to arrive on time at the same instance.

Probability Union Rule

The probability union rule is another major pillar in probability theory, vital for calculating the probability of either one event or another (or both) occurring. This is particularly useful when you need to find the likelihood of multiple events happening without counting their intersections more than once. In our exercise, this concept is used to calculate the probability of the student passing the tests, defined as either passing Test I & II or Test I & III.

The formula for the probability of the union of two events is given by:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Here, we add the probabilities of each event and subtract the probability of both events occurring together. This subtraction is crucial because, without it, the overlap where both A and B occur would be double-counted.

In practical terms, consider two separate events, like it raining this afternoon or during the evening. With the union rule, you can figure out the chance of it raining at least once during the day by recognizing where those possibilities overlap and adjusting accordingly. Understanding how to apply this rule ensures accurate predictions when dealing with real-world problems involving multiple simultaneous possibilities.