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Chapter 1: Problem 3

Let the subsets \(C_{1}=\left\\{\frac{1}{4}

Short Answer

Expert verified
\(P_{X}\left(C_{1} \cup C_{2}\right) = \frac{5}{8}, P_{X}\left(C_{1}^{c}\right) = \frac{7}{8}, P_{X}\left(C_{1}^{c} \cap C_{2}^{c}\right) = \frac{3}{8}\)

Step by step solution

01

Find \(P_{X}\left(C_{1} \cup C_{2}\right)\)

The union of two sets is comprised of elements that are in either set. In this case, as \(C_{1}\) and \(C_{2}\) are disjoint, that means they don't have common elements, we can simply add their probabilities to find the union. So, \(P_{X}\left(C_{1} \cup C_{2}\right) = P_{X}\left(C_{1}\right) + P_{X}\left(C_{2}\right) = \frac{1}{8} + \frac{1}{2} = \frac{5}{8}.
02

Find \(P_{X}\left(C_{1}^{c}\right)\)

The complement of a set \(C_{1}\) consists of all outcomes which are not in \(C_{1}\). It's denoted as \(C_{1}^{c}\). To find the probability of the complement of \(C_{1}\), we subtract the probability of \(C_{1}\) from 1, because the total probability of an event is 1. So, \(P_{X}\left(C_{1}^{c}\right) = 1 - P_{X}\left(C_{1}\right) = 1 - \frac{1}{8} = \frac{7}{8}.
03

Find \(P_{X}\left(C_{1}^{c} \cap C_{2}^{c}\right)\)

The intersection of the complements of \(C_{1}\) and \(C_{2}\), denoted as \(P_{X}\left(C_{1}^{c} \cap C_{2}^{c}\right)\), becomes the set of all outcomes not in either \(C_{1}\) or \(C_{2}\). To find this, we can use the result found in Step 2 and subtract \(P_{X}\left(C_{2}\right) from it because those are the outcomes not in \(C_{2}\) but were included in the calculation of \(P_{X}\left(C_{1}^{c}\right)\). Thus, \(P_{X}\left(C_{1}^{c} \cap C_{2}^{c}\right) = P_{X}\left(C_{1}^{c}\right) - P_{X}\left(C_{2}\right) = \frac{7}{8} - \frac{1}{2} = \frac{3}{8}.

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