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Chapter 1: Problem 15

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

Short Answer

Expert verified
In order to prove the given set equalities, we analyzed the relationships between the sets using set diagrams and their operations like union, intersection, and complement. We found that for any element in E, it must either be in \(E \cap F\) or \(E \cap F^c\), which demonstrates that \(E = E \cap F \cup E \cap F^c\). Similarly, for any element in \(E \cup F\), it must either be in E or \(F \cap E^c\), which demonstrates that \(E \cup F = E \cup F \cap E^c\). Through this analysis, we have successfully argued and shown that both set equalities hold true.

Step by step solution

01

Prove \(E = E \cap F \cup E \cap F^c\)

First, let's try to prove that \(E = E \cap F \cup E \cap F^c\). We can argue that every element in E must be in either \(E \cap F\) or \(E \cap F^c\), because these two sets encompass all possibilities for elements in E. Let's break it down further: - If an element x is in E and in F, then x is in \(E \cap F\). - If an element x is in E but not in F (x is in F's complement), then x is in \(E \cap F^c\). Since the union operation combines all elements from both sets, it is clear that \(E = E \cap F \cup E \cap F^c\).
02

Prove \(E \cup F = E \cup F \cap E^c\)

Now, let's try to prove that \(E \cup F = E \cup F \cap E^c\). We can argue that every element in \(E \cup F\) must be in either E or \(F \cap E^c\), because these two sets encompass all possibilities for elements in the union. Let's break it down further: - If an element x is in E, it is automatically in \(E \cup F\). - If an element x is in F but not in E (x is in E's complement), then x is in \(F \cap E^c\). Since the union operation combines all elements from both sets, it is clear that \(E \cup F = E \cup F \cap E^c\). Through these two steps, we have argued and demonstrated that both set equalities hold: 1. \(E = E \cap F \cup E \cap F^c\) 2. \(E \cup F = E \cup F \cap E^c\)

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Key Concepts

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

Union of Sets
The concept of a "Union of Sets" is fundamental in set theory. When we talk about the union of two sets, we refer to a new set that contains all the elements that are in either or both of the original sets. Let's denote two sets as \( A \) and \( B \). The union of these sets is represented as \( A \cup B \). This set consists of every element from \( A \) as well as every element from \( B \).
For example, if \( A = \{1, 2, 3\} \) and \( B = \{3, 4, 5\} \), the union \( A \cup B \) would be \( \{1, 2, 3, 4, 5\} \). Notice that the number 3 is present in both sets, but it appears only once in the union.
  • Each element is present only once in the union, even if it appears in both original sets.
  • The union operation is commutative, meaning that \( A \cup B = B \cup A \).
  • It is also associative, so \( A \cup (B \cup C) = (A \cup B) \cup C \).
In solving problems, the union helps us to combine elements, as seen in the exercise where \( E = E \cap F \cup E \cap F^c \), encompassing all elements in \( E \).
Intersection of Sets
The "Intersection of Sets" is another essential concept. The intersection refers to elements that are common between two sets. If we have two sets \( A \) and \( B \), their intersection is represented by \( A \cap B \). This new set includes only the elements that \( A \) and \( B \) share.
For instance, if \( A = \{1, 2, 3\} \) and \( B = \{3, 4, 5\} \), the intersection \( A \cap B \) would be \( \{3\} \). The number 3 is the only element present in both sets.
  • The intersection might result in an empty set, denoted by \( \emptyset \), if no common elements exist.
  • Much like the union, intersection is commutative: \( A \cap B = B \cap A \).
  • It is also associative: \( A \cap (B \cap C) = (A \cap B) \cap C \).
Understanding intersections helped us demonstrate in the exercise that elements in \( E \) can be divided into those also in \( F \) and those not in \( F \).
Complement of a Set
The "Complement of a Set" is about understanding what is outside a particular set within a universal set context. If \( A \) is a set, its complement, denoted as \( A^c \), consists of all elements not in \( A \) but within a larger set, often assumed to be the universe \( U \).
For example, if the universe \( U = \{1, 2, 3, 4, 5\} \) and \( A = \{2, 3\} \), then the complement \( A^c \) is \( \{1, 4, 5\} \). It includes everything in \( U \) that isn't in \( A \).
  • The complement focuses on the elements missing from the set in the context of the universal set.
  • The union of a set with its complement covers the entire universe: \( A \cup A^c = U \).
  • The intersection of a set with its complement results in an empty set: \( A \cap A^c = \emptyset \).
In the exercise, the complement told us about what was specifically outside the sets \( F \) and \( E \), assisting in dividing \( E \) and incorporating complementary events.

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Most popular questions from this chapter

Bill and George go target shooting together. Both shoot at a target at the same time. Suppose Bill hits the target with probability \(0.7\), whereas George, independently, hits the target with probability \(0.4 .\) (a) Given that exactly one shot hit the target, what is the probability that it was George's shot? (b) Given that the target is hit, what is the probability that George hit it?

For events \(E_{1}, E_{2}, \ldots, E_{n}\) show that \(P\left(E_{1} E_{2} \cdots E_{n}\right)=P\left(E_{1}\right) P\left(E_{2} \mid E_{1}\right) P\left(E_{3} \mid E_{1} E_{2}\right) \cdots P\left(E_{n} \mid E_{1} \cdots E_{n-1}\right)\)

Sixty percent of the families in a certain community own their own car, thirty percent own their own home, and twenty percent own both their own car and their own home. If a family is randomly chosen, what is the probability that this family owns a car or a house but not both?

Let \(E\) and \(F\) be mutually exclusive events in the sample space of an experiment. Suppose that the experiment is repeated until either event \(E\) or event \(F\) occurs. What does the sample space of this new super experiment look like? Show that the probability that event \(E\) occurs before event \(F\) is \(P(E) /\) \([P(E)+P(F)]\) Hint: Argue that the probability that the original experiment is performed \(n\) times and \(E\) appears on the \(n\) th time is \(P(E) \times(1-p)^{n-1}, n=1,2, \ldots\), where \(p=P(E)+P(F)\). Add these probabilities to get the desired answer.

The dice game craps is played as follows. The player throws two dice, and if the sum is seven or eleven, then she wins. If the sum is two, three, or twelve, then she loses. If the sum is anything else, then she continues throwing until she either throws that number again (in which case she wins) or she throws a seven (in which case she loses). Calculate the probability that the player wins.
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