91Ó°ÊÓ

The given information is that \(C_{1}, C_{2}, C_{3}\) are independent events, and, their individual probabilities are \(P(C_{1}) = \frac{1}{2}, P(C_{2}) = \frac{1}{3}, P(C_{3}) = \frac{1}{4}\). Compute the probability, \(P(C_{1}\cup C_{2} \cup C_{3})\), that at least one of these events happens. The idea here is to use the Principle of Inclusion and Exclusion to solve the problem.

The Principle of Inclusion and Exclusion is used when calculating the probability of the union of multiple events. The formula is represented as \(P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)\). However, considering that \(C_{1}, C_{2}, and C_{3}\) are independent events, their intersection is the product of their probabilities. Therefore, \(P(C_{1} \cap C_{2}) = P(C_{1}) \cdot P(C_{2})\), \(P(C_{1} \cap C_{3}) = P(C_{1}) \cdot P(C_{3})\), \(P(C_{2} \cap C_{3}) = P(C_{2}) \cdot P(C_{3})\), and \(P(C_{1} \cap C_{2} \cap C_{3}) = P(C_{1}) \cdot P(C_{2}) \cdot P(C_{3})\). By calculating these values and substituting them into the equation of the Principle of Inclusion and Exclusion, we can get the solution.

Substitute the calculated values of intersections and the given probabilities into the equation \(P(C_{1} \cup C_{2} \cup C_{3})= \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{2} \cdot \frac{1}{3} - \frac{1}{2} \cdot \frac{1}{4} - \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4}\). By simplifying this equation, the final answer can be obtained.