A problem is given to three students whose chances of solving it are $\frac{1}{2}$, $\frac{1}{3}$ and $\frac{1}{4}$ respectively. What is the probability that the problem will be solved?

Let $A,B,$ and $C$ be the events that the first, second, and third student solve the problem respectively.
The given probabilities are:

$$P(A)=\frac{1}{2},P(B)=\frac{1}{3},P(C)=\frac{1}{4}$$

The problem will be solved if at least one student solves it. It is easier to calculate the probability that the problem is not solved by any of them, and subtract it from $1$.

The probabilities of individual students failing to solve the problem are:

$$P(A^{\prime })=1-\frac{1}{2}=\frac{1}{2}$$

$$P(B^{\prime })=1-\frac{1}{3}=\frac{2}{3}$$

$$P(C^{\prime })=1-\frac{1}{4}=\frac{3}{4}$$

Since the events are independent, the probability that none of them solves the problem is:

$$P(\text{None solves})=P(A^{\prime })\times P(B^{\prime })\times P(C^{\prime })=\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}=\frac{1}{4}$$

Therefore, the probability that the problem will be solved is:

$$P(\text{Solved})=1-P(\text{None solves})=1-\frac{1}{4}=\frac{3}{4}$$