Probability of Solving a Statistics Problem by Independent Students

Suppose a problem in statistics is given to three students, A, B, and C, whose probabilities of solving it independently are 1/2, 1/3, and 1/4 respectively. The question is: What is the probability that the problem will be solved if all of them try independently?

Understanding the Problem

First, let's define the probabilities:

The probability that A can solve the problem is PA 1/2. The probability that A cannot solve the problem is PA' 1/2. The probability that B can solve the problem is PB 1/3. The probability that B cannot solve the problem is PB' 2/3. The probability that C can solve the problem is PC 1/4. The probability that C cannot solve the problem is PC' 3/4.

Solving the Problem

We are interested in the probability that exactly one of them will solve the problem. Let's break this down step by step:

A solves, B does not solve, and C does not solve. A does not solve, B solves, and C does not solve. A does not solve, B does not solve, and C solves.

Calculating Each Scenario

A solves, B does not solve, and C does not solve: Probability of this happening: PA * PB' * PC' (1/2) * (2/3) * (3/4) 6/24 A does not solve, B solves, and C does not solve: Probability of this happening: PA' * PB * PC' (1/2) * (1/3) * (3/4) 3/24 A does not solve, B does not solve, and C solves: Probability of this happening: PA' * PB' * PC (1/2) * (2/3) * (1/4) 2/24

Summing the Probabilities

The combined probability that exactly one of them will solve the problem is:

6/24 3/24 2/24 11/24

Therefore, the probability that the problem will be solved by exactly one of the students is 11/24.

Conclusion

The final answer to the problem is 11/24.

Additional Insights

Understanding the problem involves the concept of independent events. Each student's ability to solve the problem is independent of the others. This means that the probability of one student solving the problem does not affect the probability of the others solving or not solving the problem.

It's important to break the problem down into its constituent parts and calculate the probabilities for each scenario separately before summing them up to find the overall probability.