Probability Part 1

Given $P(A)=\;\frac{1}{2}, P(B)=\;\frac{1}{3}, P(C)=\;\frac{1}{4}$ So, $P(\overline{A})=\;\frac{1}{2}$ (Probability that the problem can't be solve by A) $P(\overline{B})=\;\frac{2}{3}$ (Probability that the problem can't be solve by B) and $P(\overline{C})=\;\frac{3}{4}$ (Probability that the problem can't be solve by $C$ ) Now the probability that the problem is solved by any one student of $A, B$ and $C=1$ - the probability that the problem is solved by none of the students of $A, B$ and $C$ $P(A \cup B \cup C)=1-P(\overline{A}) P(\overline{B}) P(\overline{C})$ $=1-\;\frac{1}{2} \times \;\frac{2}{3} \times \;\frac{3}{4}$ $=\;\frac{3}{4}$