1. Let the probabilities that students $A,B,C$ solve the problem be:

$P(A)=\frac12, \qquad P(B)=\frac13, \qquad P(C)=\frac14$

We want the probability that the problem is solved by at least one student.

2. It is easier to use the complement rule:

$P(\text{problem is solved}) = 1 - P(\text{none of them solves})$

3. Compute the probability that each student does not solve the problem:

$P(A^c)=1-\frac12=\frac12$ $P(B^c)=1-\frac13=\frac23$ $P(C^c)=1-\frac14=\frac34$

4. Assuming the students attempt independently, the probability that none solves is:

$P(A^c \cap B^c \cap C^c)=\frac12 \cdot \frac23 \cdot \frac34$

$= \frac{1\cdot 2\cdot 3}{2\cdot 3\cdot 4}=\frac14$

5. Therefore,

$P(\text{problem is solved}) = 1 - \frac14 = \frac34$

6. Compare with the options:

• A: $\frac34$ ✅
• B: $\frac12$
• C: $\frac23$
• D: $\frac13$

So the correct option is A.