Understanding the Problem: Probability of Solving a Problem

The question asks for the probability that a problem is solved by at least one of three students, A, B, and C, given their individual probabilities of solving the problem. The key here is that the students solve the problem independently.

Given Probabilities

• Probability that student A solves the problem, \(P(A) = \frac{3}{7}\)
• Probability that student B solves the problem, \(P(B) = \frac{5}{9}\)
• Probability that student C solves the problem, \(P(C) = \frac{1}{5}\)

Concept of Independent Events

Since the students solve the problem independently, the outcome of one student attempting the problem does not affect the outcome of another student attempting it. This independence is crucial for calculating combined probabilities.

Strategy: Using the Complement Rule

Calculating the probability that 'at least one student solves the problem' can be complex as it involves cases where A solves, B solves, C solves, A and B solve, A and C solve, B and C solve, or all three solve. A simpler approach is to use the complement rule. The complement of the event 'the problem is solved' is the event 'the problem is NOT solved'. The problem is NOT solved only if NONE of the students solve it.

So, \(P(\text{problem is solved}) = 1 - P(\text{problem is NOT solved})\).

\(P(\text{problem is NOT solved}) = P(\text{A does not solve AND B does not solve AND C does not solve})\).

Calculating Probabilities of Not Solving

If the probability of solving is P(X), the probability of not solving is \(P(X') = 1 - P(X)\).

• Probability that A does not solve, \(P(A') = 1 - P(A) = 1 - \frac{3}{7} = \frac{7-3}{7} = \frac{4}{7}\)
• Probability that B does not solve, \(P(B') = 1 - P(B) = 1 - \frac{5}{9} = \frac{9-5}{9} = \frac{4}{9}\)
• Probability that C does not solve, \(P(C') = 1 - P(C) = 1 - \frac{1}{5} = \frac{5-1}{5} = \frac{4}{5}\)

Calculating Probability that None Solve the Problem

Since the events are independent, the probability that none of them solve the problem is the product of their individual probabilities of not solving:

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

\(P(\text{none solve}) = \frac{4}{7} \times \frac{4}{9} \times \frac{4}{5}\)

\(P(\text{none solve}) = \frac{4 \times 4 \times 4}{7 \times 9 \times 5}\)

\(P(\text{none solve}) = \frac{64}{315}\)

Calculating Probability that the Problem is Solved

Now, we use the complement rule:

\(P(\text{problem is solved}) = 1 - P(\text{none solve})\)

\(P(\text{problem is solved}) = 1 - \frac{64}{315}\)

\(P(\text{problem is solved}) = \frac{315 - 64}{315}\)

\(P(\text{problem is solved}) = \frac{251}{315}\)

Conclusion

The probability that the problem will be solved by at least one of the three students is \(\frac{251}{315}\).

Revision Table: Probability Concepts

Additional Information: More on Probability of Multiple Events

When dealing with multiple events, understanding their relationship is key. If events are mutually exclusive (cannot happen at the same time), the probability of A or B happening is \(P(A \text{ or } B) = P(A) + P(B)\). However, in this problem, students solving the problem are independent events, not mutually exclusive (multiple students can solve it). The probability of at least one independent event happening is best found using the complement rule, as demonstrated in the solution.

For independent events A, B, C, the probability of all three happening is \(P(A \text{ and } B \text{ and } C) = P(A) \times P(B) \times P(C)\).

The problem solved here is a classic example illustrating the application of the complement rule in probability, particularly useful when dealing with independent events and the concept of "at least one".