Introduction / Context:
This is a standard probability problem about independent events and the complement rule. Three students attempt the same problem independently, each with a given probability of solving it. Instead of directly calculating all cases in which the problem is solved, the more efficient approach is to calculate the probability that none of them solves it and then subtract that value from 1. This is a common technique in probability questions involving "at least one" events.
Given Data / Assumptions:
Student 1 solves the problem with probability 1/2.
Student 2 solves the problem with probability 1/3.
Student 3 solves the problem with probability 1/4.
Their attempts are independent of each other.
Concept / Approach:The event "the problem is solved" is the complement of the event "none of the students solves the problem". For independent events, the probability that none of them succeeds is the product of the probabilities of each student failing. Once that value is found, the required probability is 1 minus this product. Using the complement avoids having to sum multiple overlapping cases where one, two or three students succeed.
Step-by-Step Solution:Let S1, S2 and S3 be the events that student 1, 2 and 3 solve the problem respectively.Given: P(S1) = 1/2, P(S2) = 1/3, P(S3) = 1/4.Therefore, probabilities that they do not solve the problem are:P(not S1) = 1 - 1/2 = 1/2.P(not S2) = 1 - 1/3 = 2/3.P(not S3) = 1 - 1/4 = 3/4.The probability that none of them solves the problem is the product of these, since the attempts are independent:P(none solves) = (1/2) * (2/3) * (3/4).Multiply: (1/2) * (2/3) = 1/3, and (1/3) * (3/4) = 1/4.So P(none solves) = 1/4.The required probability that at least one solves the problem is:P(at least one solves) = 1 - P(none solves) = 1 - 1/4 = 3/4.Verification / Alternative check:We could also expand cases where exactly one, exactly two or all three solve the problem, but that is more cumbersome.The complement method works because the only way the problem is not solved is when all three fail simultaneously.Since our final probability is between 0 and 1 and matches the intuitive idea that the combined chance should be high but not certain, it is reasonable and confirmed by the arithmetic.Why Other Options Are Wrong:A value of 1/4 corresponds to the probability that none solves it, not that at least one solves. 1/2 and 2/3 underestimate the combined success probability for three students. The option 7/12 arises from adding individual probabilities without proper inclusion exclusion and is not correct for independent events. Only 3/4 correctly represents 1 minus the probability that all three fail.
Common Pitfalls:Many students mistakenly add the individual probabilities directly, writing 1/2 + 1/3 + 1/4, which can exceed 1 and does not reflect independence or overlap. Others forget to use the complement rule and attempt to compute probabilities for each combination of successes and failures, increasing the chance of counting errors. Remembering that "at least one" is best handled via 1 minus "none" is very helpful in such questions.
Final Answer:The probability that the problem is solved by at least one student is 3/4.
Discussion & Comments