Problem 4 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

A question of mathematics is given to three students to solve. Probabilities of solving the question by them are \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\), respectively. If they try to solve it, what is the probability that the problem will be solved?

Step-by-Step Solution

Answer

The probability that the problem will be solved is \( \frac{3}{4} \).

Step 1: Understanding Individual Probabilities

Each of the three students has a certain probability of solving the problem. Let's denote these probabilities as: \( P(A) = \frac{1}{2} \) for the first student, \( P(B) = \frac{1}{3} \) for the second student, and \( P(C) = \frac{1}{4} \) for the third student. These numbers refer to their chances of successfully solving the problem individually.

Step 2: Calculating the Probability that Each Student Fails

To find the probability that each student fails to solve the problem, we need to subtract their solving probabilities from 1. Thus, the failing probabilities are: \( P(A') = 1 - \frac{1}{2} = \frac{1}{2} \), \( P(B') = 1 - \frac{1}{3} = \frac{2}{3} \), \( P(C') = 1 - \frac{1}{4} = \frac{3}{4} \).

Step 3: Finding the Probability that All Students Fail

To find the probability that the problem is not solved by anyone (all fail), we multiply their individual fail probabilities: \( P(A' \cap B' \cap C') = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{1}{4} \).

Step 4: Calculating the Probability that At Least One Solves It

The probability that the problem is solved by at least one student is the complement of all failing, given by: \( 1 - P(A' \cap B' \cap C') = 1 - \frac{1}{4} = \frac{3}{4} \). This accounts for any scenario where at least one student solves the problem.

Key Concepts

Individual Probability

In probability theory, an individual probability represents the likelihood of a single event occurring. For instance, when considering how likely each student is to solve the problem on their own, we denote their probabilities as follows:
- The first student has a probability of solving the problem: \( P(A) = \frac{1}{2} \).
- The second student's probability: \( P(B) = \frac{1}{3} \).
- The third student's probability: \( P(C) = \frac{1}{4} \).
This means that each student has a certain chance of solving the problem completely independent of the others. The probabilities must be between 0 and 1, where 0 means impossible and 1 means certain. Here, knowing these values helps us understand how likely it is for individual students to solve the problem.

Complementary Events

Complementary events are pairs of outcomes in probability where one event occurring means the other cannot occur. In this context, if a student solves the problem (event A), the complementary event (event A') is that the student does not solve it.

To determine the probability that a student doesn't solve the problem, we subtract their solving probability from 1. This calculation gives us the complementary probability, expressed mathematically as:
- For the first student \( A' \): \( P(A') = 1 - \frac{1}{2} = \frac{1}{2} \).
- For the second student \( B' \): \( P(B') = 1 - \frac{1}{3} = \frac{2}{3} \).
- For the third student \( C' \): \( P(C') = 1 - \frac{1}{4} = \frac{3}{4} \).
This concept is crucial because understanding complementary events helps us calculate scenarios where specific outcomes do not happen, which is useful for determining overall probabilities.

Cumulative Probability

Cumulative probability is the total probability that accounts for multiple potential outcomes. In this exercise, we want to find the probability that the problem gets solved by at least one student.

The key in cumulative probability is using complementary events to simplify calculations. First, we calculate the probability that all students fail to solve the problem:
- \( P(A' \cap B' \cap C') = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{1}{4} \).
This result indicates that there is a 25% chance none of the students solve the problem. The complement of this event will give us the probability that at least one student successfully solves the problem:
- \( 1 - P(A' \cap B' \cap C') = 1 - \frac{1}{4} = \frac{3}{4} \).
Therefore, the cumulative probability that at least one of the students solves the problem is \( \frac{3}{4} \), or 75%. This comprehensive probability accounts for any successful outcome combination.