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Chapter 1: Problem 15

Argue that \(E=E F \cup E F^{c}, E \cup F=E \cup F E^{e}\).

Short Answer

Expert verified
To prove the identities \(E = EF \cup EF^c\) and \(E \cup F = E \cup FE^c\), we can show that every element in one set is also an element in the other set and vice versa, demonstrating that the sets are subsets of each other. By proving \(E \subseteq EF \cup EF^c\) and \(EF \cup EF^c \subseteq E\), we conclude that \(E = EF \cup EF^c\). Similarly, by proving \(E \cup F \subseteq E \cup FE^c\) and \(E \cup FE^c \subseteq E \cup F\), we conclude that \(E \cup F = E \cup FE^c\).

Step by step solution

01

Prove that \(E = EF \cup EF^c\)

To prove that \(E = EF \cup EF^c\), we need to show that every element in \(E\) is also an element in either \(EF\) or \(EF^c\), and vice versa. First, let's take an element \(x \in E\). Since \(x \in E\), either \(x \in F\) or \(x \in F^c\). If \(x \in F\), then \(x \in EF\) because \(EF\) contains all the elements that are in both \(E\) and \(F\). If \(x \in F^c\), then \(x \in EF^c\) for the same reason. Thus, every element in \(E\) is also an element in either \(EF\) or \(EF^c\), which means \(E \subseteq EF \cup EF^c\). Now, let's take an element \(x \in EF \cup EF^c\). If \(x \in EF\), then \(x \in E\) because \(EF\) is the intersection of \(E\) and \(F\). If \(x \in EF^c\), then \(x \in E\) because \(EF^c\) is the intersection of \(E\) and \(F^c\). This shows that every element in \(EF \cup EF^c\) is also an element in \(E\), which means \(EF \cup EF^c \subseteq E\). Since we have shown that \(E \subseteq EF \cup EF^c\) and \(EF \cup EF^c \subseteq E\), we can conclude that \(E = EF \cup EF^c\).
02

Prove that \(E \cup F = E \cup FE^c\)

To prove that \(E \cup F = E \cup FE^c\), we need to show that every element in \(E \cup F\) is also an element in \(E \cup FE^c\), and vice versa. First, let's take an element \(x \in E \cup F\). If \(x \in E\), then \(x \in E \cup FE^c\) because it's already included in the first set. If \(x \in F\), then either \(x \in E\) or \(x \in E^c\). If \(x \in E\), \(x\) is already in \(E \cup FE^c\). If \(x \in E^c\), then \(x \in FE^c\), and thus \(x \in E \cup FE^c\). Therefore, every element in \(E \cup F\) is also an element in \(E \cup FE^c\), which means \(E \cup F \subseteq E \cup FE^c\). Now, let's take an element \(x \in E \cup FE^c\). If \(x \in E\), then \(x \in E \cup F\) because it's already included in the first set. If \(x \in FE^c\), then \(x \in F\) but \(x \notin E\). Since \(x \in F\), \(x \in E \cup F\). This shows that every element in \(E \cup FE^c\) is also an element in \(E \cup F\), which means \(E \cup FE^c \subseteq E \cup F\). Since we have shown that \(E \cup F \subseteq E \cup FE^c\) and \(E \cup FE^c \subseteq E \cup F\), we can conclude that \(E \cup F = E \cup FE^c\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Union and Intersection of Sets
Understanding the union and intersection of sets is essential for delving into probability and set theory. The union of two sets, denoted as
\( A \text{ union } B \text{ or } A \cup B \), includes all the elements that are in either set A or set B, or in both. Imagine combining two circles in a Venn diagram; the area covering either circle represents their union.

Conversely, the intersection, denoted as
\( A \text{ intersection } B \text{ or } A \cap B \), consists of elements that are common to both sets. Going back to our Venn diagram analogy, this is the area where both circles overlap.

These fundamental operations allow us to understand how different scenarios can combine or overlap. For instance, in the given textbook problem, when we say \(E = EF \text{ union } EF^c\), we're asserting that all elements of set E are either in the intersection set with F or with its complement, which collectively exhaust all possibilities within set E.
Complements of Sets
The complement of a set is a concept that provides a way to express elements not in the original set, with respect to a universal set. In set theory, the complement of set A, denoted as
\( A^c \), includes all the elements that are not in set A. When it pertains to probability, the complement helps us determine the likelihood of all other outcomes not mentioned in the original set.

In considering \( EF \cup EF^c \), we're essentially saying that these two sets, the intersection of E with F and the intersection of E with the complement of F, together include every possible element from the universal set related to E. This is crucial because it allows us to cover all bases when considering the elements in set E, especially when calculating probabilities, ensuring that no potential outcome is ignored.
Elementary Probability Concepts
Elementary probability involves the basics of measuring the likelihood of events within a given set. Probability values range between 0 (an event that will not occur) and 1 (an event that is certain to happen). It's the foundation upon which more complex probability theories are built.

Probabilities are often determined using set operations like unions, intersections, and complements. For instance, the probability of either event A or B happening is calculated using the union of sets, while the probability of both events occurring simultaneously involves their intersection.

In the context of our exercise, understanding these set operations helps in determining the probability of events involving sets E and F, and their various combinations. When you grasp these elementary concepts thoroughly, you'll be better equipped to tackle more advanced problems in probability and statistics.

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Most popular questions from this chapter

If the occurrence of \(B\) makes \(A\) more likely, does the occurrence of \(A\) make \(B\) more likely?

Suppose all \(n\) men at a party throw their hats in the center of the room. Each man then randomly selects a hat. Show that the probability that none of the \(n\) men selects his own hat is $$ \frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-+\cdots \frac{(-1)^{n}}{n !} $$ Note that as \(n \rightarrow \infty\) this converges to \(e^{-1}\). Is this surprising?

Show that $$ P\left(\bigcup_{i=1}^{n} E_{i}\right) \leq \sum_{i=1}^{n} P\left(E_{i}\right) $$ This is known as Boole's inequality. Either use Equation \((1.2)\) and methematical induction, or else show that \(\bigcup_{i-1}^{n} E_{i}=\bigcup_{i=1}^{e} F_{i}\), where \(F_{1}=E_{1}, F_{i}=E_{i} \Pi_{j=1}^{l-1} E_{j}^{e}\), and use property (iii) of a probability.

An individual uses the following gambling system at Las Vegas. He bets \(\$ 1\) that the roulette wheel will come up red. If he wins, he quits. If he loses then he makes the same bet a second time only this time he bets \(\$ 2 ;\) and then regardless of the outcome, quits. Assuming that he has a probability of I of winning each bet, what is the probability that he goes home a winner? Why is this system not used by everyone?

Suppose each of three persons tosses a coin. If the outcome of one of the tosses differs from the other outcomes, then the game ends. If not, then the persons start over and retoss their coins. Assuming fair coins, what is the probability that the game will end with the first round of tosses? If all three coins are biased and have a probability \(\frac{1}{4}\) of landing heads, then what is the probability that the game will end at the first round?
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