Question:

Step 1: Identify that the problem is solved if AT LEAST ONE student solves it. Step 2: Use complement: $P($at least one solves$) = 1 - P($none solve$)$. Step 3: Find $P($each student fails$)$: $$ \begin{aligned} P(A \text{ fails}) &= 1 - \frac{1}{2} = \frac{1}{2} \\ P(B \text{ fails}) &= 1 - \frac{1}{3} = \frac{2}{3} \\ P(C \text{ fails}) &= 1 - \frac{1}{4} = \frac{3}{4} \end{aligned} $$ Step 4: Since attempts are independent: $$ \begin{aligned} P(\text{none solve}) &= P(A \text{ fails}) \times P(B \text{ fails}) \times P(C \text{ fails}) \\ &= \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{6}{24} = \frac{1}{4} \end{aligned} $$ Step 5: Therefore: $$ \begin{aligned} P(\text{problem solved}) &= 1 - \frac{1}{4} = \frac{3}{4} \end{aligned} $$ Therefore: $\frac{3}{4}$ <div class="key-concept"><strong>Key Concept:</strong> Use the complement approach: find the probability that ALL students fail, then subtract from 1. This avoids tedious case-by-case analysis when multiple people can solve the problem.</div> <div class="trap-box"><strong>Trap:</strong> Students attempt to add individual probabilities P(A) + P(B) + P(C) = 1/2 + 1/3 + 1/4, which violates the addition rule for non-mutually exclusive events. Others try to sum all possible cases (exactly one solves, exactly two solve, all three solve) and make arithmetic errors.</div>