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Chapter 1: Problem 23

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

Expert verified
The probability of the union of the given independent events \(C_{1}, C_{2}, C_{3}\) is \(\frac{25}{24}\), which is more than 1. This suggests that there is absolute certainty that at least one of the three events will occur. However, the value being greater than 1 indicates that there might be an overlap and some events could occur simultaneously.

Step by step solution

01

Statement of formula

For independent events, the probability of a union can be calculated using the following formula: \(P(A\cup B\cup C) = P(A) + P(B) + P(C) - P(A\cap B) - P(A\cap C) - P(B\cap C) + P(A\cap B\cap C)\). Since the events are independent, intersection of any two or all three is a multiplication of their individual probabilities.
02

Value Assignment

In the given problem, the events \(C_{1}, C_{2}, C_{3}\) are given to be independent and their probabilities are given as \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\) respectively.
03

Calculate the Probabilities of Intersections

Calculate the probabilities of the intersections of two events. These are \(P(C_{1}\cap C_{2}) = P(C_{1}) \cdot P(C_{2}) = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}\), \(P(C_{1}\cap C_{3}) = P(C_{1}) \cdot P(C_{3}) = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}\), \(P(C_{2}\cap C_{3}) = P(C_{2}) \cdot P(C_{3}) = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}\). The probability of the intersection of all three events is \(P(C_{1}\cap C_{2}\cap C_{3}) = P(C_{1}) \cdot P(C_{2}) \cdot P(C_{3}) = \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{24}\).
04

Compute the Probability

Substitute the values calculated in steps 2 and 3 into the formula stated in step 1. The probability of the union of the three events is \(P(C_{1}\cup C_{2}\cup C_{3}) = P(C_{1}) + P(C_{2}) + P(C_{3}) - P(C_{1}\cap C_{2}) - P(C_{1}\cap C_{3}) - P(C_{2}\cap C_{3}) + P(C_{1}\cap C_{2}\cap C_{3}) = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{6} - \frac{1}{8} - \frac{1}{12} + \frac{1}{24} = \frac{25}{24}\). This method gives us the probability of union of the three given events.

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