Let's break down this probability question step by step. We are given the individual probabilities that three students, A, B, and C, will solve a question independently. We need to find the probability that the question will be solved by at least one of them.

Understanding the Probability Problem

We have three independent events: Student A solving the question, Student B solving the question, and Student C solving the question. The question is solved if any one of them, any two of them, or all three of them solve it. Calculating all these possibilities directly can be complicated.

A simpler approach is to find the probability that the question is *not* solved at all, and then subtract this probability from 1. The question is not solved only if *none* of the students solve it.

Given Probabilities

We are given the probabilities that each student solves the question:

• Probability that Student A solves the question, \(P(A) = \frac{1}{2}\)
• Probability that Student B solves the question, \(P(B) = \frac{1}{3}\)
• Probability that Student C solves the question, \(P(C) = \frac{1}{4}\)

Calculating Probabilities of Not Solving

Since the probability of an event happening plus the probability of the event not happening is 1, we can find the probability that each student *does not* solve the question:

• Probability that Student A does not solve the question, \(P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2}\)
• Probability that Student B does not solve the question, \(P(B') = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3}\)
• Probability that Student C does not solve the question, \(P(C') = 1 - P(C) = 1 - \frac{1}{4} = \frac{3}{4}\)

Finding the Probability That None Solve

The students solving the question are independent events. Therefore, the event of one student not solving is also independent of the others not solving. To find the probability that *none* of the students solve the question, we multiply their individual probabilities of not solving:

\(P(\text{None solve}) = P(A') \times P(B') \times P(C')\)

\(P(\text{None solve}) = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}\)

Let's calculate this value:

\(P(\text{None solve}) = \frac{1 \times 2 \times 3}{2 \times 3 \times 4} = \frac{6}{24}\)

Simplifying the fraction:

\(P(\text{None solve}) = \frac{1}{4}\)

So, the probability that none of the students solve the question is \(\frac{1}{4}\).

Calculating the Probability That the Question Is Solved

The question is solved if at least one student solves it. This is the complement of the event that none of the students solve the question. Therefore, the probability that the question is solved is:

\(P(\text{Question is solved}) = 1 - P(\text{None solve})\)

\(P(\text{Question is solved}) = 1 - \frac{1}{4}\)

\(P(\text{Question is solved}) = \frac{4}{4} - \frac{1}{4} = \frac{3}{4}\)

Thus, the probability that the question will be solved is \(\frac{3}{4}\).

Probability none solve = \(P(A') \times P(B') \times P(C') = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{6}{24} = \frac{1}{4}\)

Probability at least one solves = \(1 - P(\text{none solve}) = 1 - \frac{1}{4} = \frac{3}{4}\)

Revision Table: Probability Concepts

Additional Information on Independent Probability

In probability, events are considered independent if the outcome of one event does not influence the outcome of another. In this question, the assumption is that Student A solving the question does not change the chances of Student B or Student C solving it, and vice versa. This is crucial because it allows us to multiply the probabilities of the individual events (or their complements) to find the probability of all of them happening together.

If the events were dependent, we would need additional information about how the probability of one student solving changes based on whether another student has solved it or not. For example, if they were working together and one student's success helped the others, the events would be dependent.

The method used here (calculating the complement) is very common and efficient when dealing with the probability of "at least one" occurrence among several independent events.