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Chapter 1: Problem 25

Let the three mutually independent events \(C_{1}, C_{2}\), and \(C_{3}\) be such that \(P\left(C_{1}\right)=P\left(C_{2}\right)=P\left(C_{3}\right)=\frac{1}{4} .\) Find \(P\left[\left(C_{1}^{c} \cap C_{2}^{c}\right) \cup C_{3}\right]\)

Short Answer

Expert verified
The probability \(P[(C_{1}^{c} \cap C_{2}^{c}) \cup C_{3}]\) is \(\frac{45}{64}\).

Step by step solution

01

Understand the Events and their Probabilities

Firstly, we need to understand the events and their probabilities. The exercise has specified three independent events, \(C_{1}, C_{2}\), and \(C_{3}\), each with a probability of \(\frac{1}{4}\). The notation \(C_{i}^{c}\) denotes the complement of event \(C_{i}\), meaning that the event \(C_{i}\) does not occur. The probability of these not happening given as \(P(C_{i}^{c}) = 1 - P(C_{i}) = 1 - \frac{1}{4} = \frac{3}{4}\).
02

Find Probabilities of Intersection of Events

Now we need to find the probability of the intersection of events \(C_{1}^{c}\) and \(C_{2}^{c}\). As the events are independent, the intersection of \(C_{1}^{c}\) and \(C_{2}^{c}\) is given by \( P(C_{1}^{c} \cap C_{2}^{c}) = P(C_{1}^{c}) * P(C_{2}^{c}) = \frac{3}{4} * \frac{3}{4} = \frac{9}{16}\).
03

Implementing Union of Events

The task requires us to find the probability of the union of events \((C_{1}^{c} \cap C_{2}^{c})\) and \(C_{3}\). The probability of a union of two events A and B is given by \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). Applying this to our exercise, we get:\[ P[(C_{1}^{c} \cap C_{2}^{c}) \cup C_{3}] = P(C_{1}^{c} \cap C_{2}^{c}) + P(C_{3}) - P((C_{1}^{c} \cap C_{2}^{c}) \cap C_{3}) \] Since the events are independent, the intersection \(P((C_{1}^{c} \cap C_{2}^{c}) \cap C_{3})\) can be found as \(P(C_{1}^{c} \cap C_{2}^{c}) * P(C_{3})\).
04

Calculating the Final Probability

Inputting the calculated probabilities into the union formula from step 3, the final answer is given by:\[ P[(C_{1}^{c} \cap C_{2}^{c}) \cup C_{3}] = \frac{9}{16} + \frac{1}{4} - (\frac{9}{16} * \frac{1}{4}) = \frac{9}{16} + \frac{1}{4} - \frac{9}{64} = \frac{45}{64}\].

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