Probabilities to solve a specific problem by A, B and C are $\frac{1}{2},\frac{1}{3}$ and $\frac{1}{4}$ respectively. Probability that at least one will solve the problem is:

$\frac{3}{4}$

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

P(atleast one solves the problem) = $P(A∪B∪C)$

$=P(A)+P(B)+P(C)-P(A∩B)-P(B∩C)-P(A∩C)+P(A∩B∩C)$

for independent events $(P(X∩Y∩Z)=P(X)P(Y)P(Z)$

so

$⇒P(A)+P(B)+P(C)-P(A)P(B)-P(B)P(C)-P(C)P(A)+P(A)P(B)P(C)$

$=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{2×3}-\frac{1}{3×4}-\frac{1}{2×4}+\frac{1}{2×3×4}$

$=\frac{12+8+6-4-2-3+1}{24}=\frac{21-3}{24}=\frac{18}{24}=\frac{3}{4}$