A problem is given to three students A, B and C whose probabilities of solving the problem are \(\frac{1}{2},\frac{3}{4}\) and \(\frac{1}{4}\) respectively. What is the probability that the problem will be solved if they all solve the problem independently?
Calculating Probability: Students Solving a Problem Independently
The problem asks for the probability that a given problem is solved by at least one of three students, A, B, and C, who attempt to solve it independently. We are given the individual probabilities of each student solving the problem.
Let:
\(P(A)\) be the probability that student A solves the problem.
\(P(B)\) be the probability that student B solves the problem.
\(P(C)\) be the probability that student C solves the problem.
We are given:
\(P(A) = \frac{1}{2}\)
\(P(B) = \frac{3}{4}\)
\(P(C) = \frac{1}{4}\)
The students solve the problem independently. This means the outcome of one student attempting to solve the problem does not affect the outcome of the others.
Understanding Independent Events and Probability
For independent events, the probability that all of them occur is the product of their individual probabilities. The probability that none of them occur is also the product of the probabilities of each event not occurring.
Strategy to Find the Probability the Problem is Solved
It's easier to calculate the probability that the problem is not solved by any of the students, and then subtract this from 1. The problem is not solved if and only if student A fails, student B fails, and student C fails.
Let \(P(A')\) be the probability that student A does not solve the problem, \(P(B')\) be the probability that student B does not solve the problem, and \(P(C')\) be the probability that student C does not solve the problem.
The probability of not solving is \(1\) minus the probability of solving:
\(P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2}\)
\(P(B') = 1 - P(B) = 1 - \frac{3}{4} = \frac{1}{4}\)
\(P(C') = 1 - P(C) = 1 - \frac{1}{4} = \frac{3}{4}\)
Since the events are independent, the probability that none of the students solve the problem is the product of their individual probabilities of not solving:
\(P(\text{None solve}) = P(A') \times P(B') \times P(C')\)
\(P(\text{None solve}) = \frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}\)
\(P(\text{None solve}) = \frac{1 \times 1 \times 3}{2 \times 4 \times 4}\)
\(P(\text{None solve}) = \frac{3}{32}\)
The probability that the problem is solved is \(1\) minus the probability that none of them solve it:
\(P(\text{Problem is solved}) = 1 - P(\text{None solve})\)
\(P(\text{Problem is solved}) = 1 - \frac{3}{32}\)
\(P(\text{Problem is solved}) = \frac{32}{32} - \frac{3}{32}\)
\(P(\text{Problem is solved}) = \frac{32 - 3}{32}\)
\(P(\text{Problem is solved}) = \frac{29}{32}\)
Final Probability Calculation
The probability that the problem will be solved if all three students attempt it independently is \(\frac{29}{32}\).
Let's summarize the probabilities:
| Student | Probability of Solving | Probability of Not Solving |
|---|---|---|
| A | \(\frac{1}{2}\) | \(\frac{1}{2}\) |
| B | \(\frac{3}{4}\) | \(\frac{1}{4}\) |
| C | \(\frac{1}{4}\) | \(\frac{3}{4}\) |
Probability that none solve: \(\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4} = \frac{3}{32}\)
Probability that at least one solves: \(1 - \frac{3}{32} = \frac{29}{32}\)
Revision Table: Probability Concepts
| Concept | Description | Formula Example |
|---|---|---|
| Probability of an Event | The likelihood of an event occurring. \(P(E)\) | \(0 \le P(E) \le 1\) |
| Complement Rule | The probability of an event not occurring is 1 minus the probability of it occurring. | \(P(E') = 1 - P(E)\) |
| Independent Events | Events where the outcome of one does not affect the outcome of others. | \(P(A \text{ and } B) = P(A) \times P(B)\) |
| Probability of At Least One | The probability that at least one event occurs among independent events. Calculated as 1 minus the probability that none of the events occur. | \(P(\text{at least one}) = 1 - P(\text{none})\) |
Additional Information: Solving Probability Problems
When dealing with multiple independent events and you need the probability of "at least one" event occurring, the complement rule is often the simplest approach. Calculating the probability of "none" occurring and subtracting from 1 avoids listing all possible scenarios where at least one student solves the problem (A solves, B solves, C solves, A and B solve, A and C solve, B and C solve, A and B and C solve), which can be complex.
The independence assumption is crucial. If the students' abilities to solve the problem were related (e.g., they studied together or used the same flawed resource), their attempts would not be independent, and a different approach would be needed.
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