Question

Let \(C_{1}, C_{2}, C_{3}\) be independent events with probabilities \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\), respectively. Compute \(P\left(C_{1} \cup C_{2} \cup C_{3}\right)\).

Short answer

The probability of the union of the three events \(C_{1}\), \(C_{2}\), and \(C_{3}\) is \(\frac{3}{4}\).

Step-by-step solution

Note Each Event Probabilities

We have three independent events given: \(C_{1}\) with probability \(\frac{1}{2}\), \(C_{2}\) with probability \(\frac{1}{3}\), and \(C_{3}\) with probability \(\frac{1}{4}\).

Use the Principle of Inclusion-Exclusion

The Principle of Inclusion-Exclusion states that for three events \(A\), \(B\), and \(C\), the probability of their union is given by \(P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(A ∩ C) – P(B ∩ C) + P(A ∩ B ∩ C)\). Here, since the events are independent, the intersection of any two or all three events can be expressed as a product of their respective probabilities.

Compute the Probabilities of Intersections

Calculate \( P(C_{1} ∩ C_{2}) = P(C_{1}) \times P(C_{2}) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\). Similarly, compute \(P(C_{1} ∩ C_{3}) = P(C_{1}) \times P(C_{3}) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}\), and \(P(C_{2} ∩ C_{3}) = P(C_{2}) \times P(C_{3}) = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}\). For all three, \(P(C_{1} ∩ C_{2} ∩ C_{3}) = P(C_{1}) \times P(C_{2}) \times P(C_{3}) = \frac{1}{2} \times \frac{1}{3} \times \frac{1}{4} = \frac{1}{24}\).

Put the Calculated Values to Principle of Inclusion-Exclusion Formula

Substitute the calculated values into the formula, \(P(C_{1} ∪ C_{2} ∪ C_{3}) = P(C_{1}) + P(C_{2}) + P(C_{3}) – P(C_{1} ∩ C_{2}) – P(C_{1} ∩ C_{3}) – P(C_{2} ∩ C_{3}) + P(C_{1} ∩ C_{2} ∩ C_{3}) = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{6} - \frac{1}{8} - \frac{1}{12} + \frac{1}{24}.\)

Simplify the Formula to Get the Result

Simplify the above equation to get the result. The LCD is 24. Express each term over the common denominator, and then sum and simplify: \(= \frac{12}{24} + \frac{8}{24} + \frac{6}{24} - \frac{4}{24} - \frac{3}{24} - \frac{2}{24} + \frac{1}{24} = \frac{18}{24} = \frac{3}{4}\).