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

Step-by-Step Solution

Answer

Option A: \(p=1, q=\frac{1}{2}\) is correct.

Step 1: Understand the Expression for Probability of Success

For the student to be successful, they must pass either test I and II or test I and III. The probability for this event can be expressed in terms of the intersection of events. Let event A be passing test I, event B be passing test II, and event C be passing test III. The student is successful if the intersection of the events \((A \cap B) \cup (A \cap C)\) occurs.

Step 2: Apply the Addition Rule for Probability

The expression \((A \cap B) \cup (A \cap C)\) can be rewritten using the probability addition rule: \[P((A \cap B) \cup (A \cap C)) = P(A \cap B) + P(A \cap C) - P(A \cap B \cap C) \]This rule accounts for the overlap between the two intersections (test I passing with test II and test I passing with test III).

Step 3: Substitute Given Values into the Expression

Now, substitute the given probabilities: - \(P(A) = p\)- \(P(B) = q\)- \(P(C) = \frac{1}{2}\)The expression becomes: \[P((A \cap B) \cup (A \cap C)) = pq + \frac{p}{2} - pq \times \frac{1}{2} = \frac{1}{2}\].

Step 4: Simplify the Expression

Simplify the expression: \[pq + \frac{p}{2} - \frac{pq}{2} = \frac{1}{2}\].This simplifies to:\[\frac{3pq}{2} + \frac{p}{2} = \frac{1}{2}\].

Step 5: Solve for p and q

Further simplify and solve the equation for \(p\) and \(q\):\[\frac{p(3q+1)}{2} = \frac{1}{2}\].Multiplying both sides by 2 gives:\[p(3q+1) = 1\].Check the possibilities from the options.

Step 6: Evaluate the Given Options

Check each option:- For option (A): \(p = 1\) and \(q = \frac{1}{2}\). Plug into \(1(3\cdot\frac{1}{2} + 1) = 1\) which holds true.- For other options, plug into \(p(3q+1)\) and check if they yield 1.

Step 7: Conclude Based on Calculations

Only option (A) satisfies the condition \(p(3q+1) = 1\). Options (B), (C), and (D) do not work. Hence, the answer is option (A) \(p=1, q=\frac{1}{2}\).

Key Concepts

Conditional Probability

Conditional probability is a way to figure out the likelihood of an event happening, given that another event has already occurred. This means we are not looking at the event in isolation. Instead, we're considering information we have about another event.
When we say "conditional," it is like having a condition or a restriction that modifies the normal probability scenario.
A simple example could be determining the probability of seeing someone holding an umbrella, given that it is already raining.
Mathematically, the conditional probability of event B given event A is represented as \( P(B|A) \) and is calculated using the formula:

• \( P(B|A) = \frac{P(A \cap B)}{P(A)} \)

Here, \( P(A \cap B) \) represents the probability of both A and B happening together, and \( P(A) \) is the probability of event A.
In the context of the problem, thinking about passing either a combination of tests given passing another test involves conditional probabilities.

Addition Rule for Probability

The addition rule for probability is useful when we want to find out the probability that either one of several events happens.
It can be used to calculate the union of two events, meaning that either one event occurs, or the other event occurs, or both occur together.
By using this rule, we make sure not to double-count the probability of events that might overlap.
The addition rule for two events A and B is:

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

This rule takes into consideration any intersection (commonality) the events may have to avoid miscalculating the likelihood.
For example, if we want to find the probability a student passes either test I and II or test I and III, we apply this rule.
The given solution rewrites the event's probability using the addition rule and considers any overlap between these test scenarios. This is important to conclude correctly.

Intersection of Events

Intersection of events is a concept used to describe the probability that multiple events occur at the same time. It specifically relates to the events happening together.
In probability, the intersection of two events A and B is denoted as \( A \cap B \) and represents the scenario where both A and B happen.
If we imagine a Venn diagram, the intersection is the part where both circles overlap.
Mathematically, the probability of the intersection of A and B is given by \( P(A \cap B) \).In the example problem, when discussing passing tests, the intersection is crucial. For a student to pass both test I and test II simultaneously, we calculate the intersection of these events.
It's the same when considering the intersection of test I and II or test I and III. Handling intersections ensures all shared outcomes are correctly accounted for, making them essential for solving probability problems accurately.