Problem 7 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

The probability of student \(A\) passing an examination is \(3 / 7\) and of student \(B\) passing is \(5 / 7\). Assuming the two events ' \(A\) passes', ' \(B\) passes', as independent, find the probability of: (i) only a passing the examination. (ii) only one of them passing the examination.

Step-by-Step Solution

Answer

(i) \(\frac{6}{49}\); (ii) \(\frac{26}{49}\).

Step 1: Understand the Problem

Read the given probabilities: Probability of student \( A \) passing is \( \frac{3}{7} \) and probability of student \( B \) passing is \( \frac{5}{7} \). The events are independent.

Step 2: Find the Probability of A Passing and B Failing (i)

The probability of only \( A \) passing is when \( A \) passes and \( B \) fails. Calculate it as: \( P(A) \cdot P(B^c) \). The probability that \( B \) fails is \( 1 - P(B) = 1 - \frac{5}{7} = \frac{2}{7} \). Then, \( P(A \text{ and } B^c) = \frac{3}{7} \times \frac{2}{7} = \frac{6}{49} \).

Step 3: Find the Probability of B Passing and A Failing (ii)

The probability of only \( B \) passing is when \( B \) passes and \( A \) fails. Calculate it as: \( P(B) \cdot P(A^c) \). The probability that \( A \) fails is \( 1 - P(A) = 1 - \frac{3}{7} = \frac{4}{7} \). Then, \( P(B \text{ and } A^c) = \frac{5}{7} \times \frac{4}{7} = \frac{20}{49} \).

Step 4: Calculate Total Probability of Only One Passing (ii)

To find the probability that only one of them passes the examination, add the probabilities found in Steps 2 and 3: \( P(A \text{ and } B^c) + P(B \text{ and } A^c) = \frac{6}{49} + \frac{20}{49} = \frac{26}{49} \).

Key Concepts

Independent Events

Understanding independent events is crucial in probability theory. Independent events are those whose occurrence does not affect the probability of the other event occurring. In simple terms, knowing whether one event has occurred does not change the likelihood of another event happening.

• Example: In our exercise, the passing of student A and the passing of student B are independent events. This means that the chances of A passing do not impact the chances of B passing and vice versa.
• Mathematically, if events A and B are independent, the probability that both events occur is the product of their individual probabilities: \[P(A \text{ and } B) = P(A) \cdot P(B)\]

By assuming independence, we could perform simple calculations to find the probability of only one student passing by multiplying probabilities.

Conditional Probability

Conditional probability is a way of finding the probability of an event occurring given that another event has already occurred. While we did not directly use conditional probabilities in the exercise, it's related to concepts like independence.When two events are independent, the occurrence of one event does not change the probability of the other. Hence, the conditional probability can be stated as:

• For independent events: \[P(A | B) = P(A)\]
• For event A given B (and the reverse): \[P(B | A) = P(B)\]

These equations highlight that, in independent events, the probability of one given the other remains unchanged. This simplifies calculations in probability problems when independence is stated.

Complementary Probability

Complementary probability helps us determine the likelihood of an event not occurring. It is determined by subtracting the probability of the event occurring from 1. Recognizing complementary probabilities is essential to solve many probability questions, such as finding the probability of an event's opposite outcome.

• In our exercise, to find the probability of student B failing, we used complementary probability: \[P(B^c) = 1 - P(B)\]
• This gave us: \[P(B^c) = 1 - \frac{5}{7} = \frac{2}{7}\]

Complementary probability is a powerful tool in probability theory, allowing us to consider non-occurrences to find other probabilities, such as in the calculations for 'only one passing' outcomes.