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Let \(A_{1}, A_{2}, \ldots, A_{k}\) be any set of events defined on a sample space \(S\). What outcomes belong to the event $$ \left(A_{1} \cup A_{2} \cup \cdots \cup A_{k}\right) \cup\left(A_{1}^{C} \cap A_{2}^{C} \cap \cdots \cap A_{k}^{C}\right) $$

Short Answer

Expert verified
The outcomes that belong to this event are all the outcomes in the sample space \(S\). This is because we are uniting the set with all outcomes included in any \(A_i\) event and the set with all outcomes not included in any \(A_i\) event, thus covering all possibilities within the sample space.

Step by step solution

01

Understand the terms

Let's start by understanding what each term represents. \(A_{1} \cup A_{2} \cup \cdots \cup A_{k}\) is the set that includes all outcomes that occur in any of the events \(A_i\). On the other hand, \(A_{1}^{C} \cap A_{2}^{C} \cap \cdots \cap A_{k}^{C}\) is the set of outcomes that do not belong to any of the events.
02

Understand the operations

A Union operation '\(\cup\)' between two sets produces a new set that includes all the elements that are in either set. An Intersection operation '\(\cap\)' produces a new set that includes only the elements that are in both sets. The Complement operation '\(^C\)' of a set includes all outcomes that are not in the given set.
03

Combine the outcomes

In the expression \(\left(A_{1} \cup A_{2} \cup \cdots \cup A_{k}\right) \cup\left(A_{1}^{C} \cap A_{2}^{C} \cap \cdots \cap A_{k}^{C}\right)\), you are basically uniting all outcomes that are in any of the \(A_i\) events with the ones that are not in any \(A_i\) events. Therefore, the final set includes any possible outcome in the sample space \(S\).

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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
In set theory, understanding the union and intersection operations is fundamental. These operations allow us to combine sets in different ways to form new sets. A union of sets, denoted by the symbol \( \cup \), includes all elements that are present in at least one of the sets. For example, if you have two sets, \( A = \{1, 2, 3\} \) and \( B = \{3, 4, 5\} \), the union \( A \cup B \) would be \( \{1, 2, 3, 4, 5\} \). On the other hand, the intersection of sets, denoted by \( \cap \), includes only those elements that are present in both sets. Using our previous example, the intersection \( A \cap B \) would be \( \{3\} \) because 3 is the only element common to both sets. You can also consider more than two sets in these operations. For instance, the union \( A_1 \cup A_2 \cup \cdots \cup A_k \) includes any element found in at least one of the sets \( A_1, A_2, \ldots, A_k \). Similarly, the intersection \( A_1 \cap A_2 \cap \cdots \cap A_k \) includes only those elements found in every set. These operations form the basics of organizing outcomes in probability or kin phenomena in mathematics, enabling a structured way to deal with complex scenarios.
Complement of a Set
The complement of a set is another crucial concept in set theory. The complement of a set \( A \), denoted by \( A^C \), consists of all the elements not present in \( A \), but in the universal set or sample space. If we're considering a sample space \( S \), then \( A^C = S \setminus A \). It’s effectively the opposite or the remainder of the sample space.Consider a situation where your sample space \( S \) contains all the numbers from 1 to 10, \( S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \), and your set \( A = \{1, 2, 3, 4, 5\} \). The complement \( A^C \) would be \( \{6, 7, 8, 9, 10\} \), as these are the elements not in \( A \).Why is this important? Complements allow us to explore scenarios where we consider everything outside of a particular set. In probability, for instance, if you’re calculating the probability of not rolling a certain number in a dice game, you’re effectively considering the complement of the set containing the number rolled.
Sample Space in Probability
The concept of a sample space is foundational in probability theory. The sample space, typically denoted by \( S \), is the set of all possible outcomes of a random experiment. If you were rolling a six-sided die, for example, the sample space would be \( S = \{1, 2, 3, 4, 5, 6\} \), as these are all the possible outcomes.Understanding sample spaces helps in clearly defining what outcomes are possible in any given scenario. Once the sample space is established, you can start working with sets of outcomes, which are called events in probability. Events are subsets of the sample space. For instance, getting an even number when rolling a die is an event which can be represented by the set \( E = \{2, 4, 6\} \). Events allow for calculation of probabilities. If each outcome in \( S \) is equally likely, the probability of an event \( E \) is calculated by dividing the number of favorable outcomes by the total number of outcomes in \( S \).This structured approach ensures clarity when calculating probabilities and enables more complex analyses when multiple events are involved.

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Most popular questions from this chapter

Suppose that \(P(A)=\frac{1}{4}\) and \(P(B)=\frac{1}{8}\). (a) What does \(P(A \cup B)\) equal if 1\. \(A\) and \(B\) are mutually exclusive? 2\. \(A\) and \(B\) are independent? (b) What does \(P(A \mid B)\) equal if 1\. \(A\) and \(B\) are mutually exclusive? 2\. \(A\) and \(B\) are independent?

An urn contains \(w\) white chips, \(b\) black chips, and \(r\) red chips. The chips are drawn out at random, one at a time, with replacement. What is the probability that a white appears before a red?

Your favorite college football team has had a good season so far but they need to win at least two of their last four games to qualify for a New Year's Day bowl bid. Oddsmakers estimate the team's probabilities of winning each of their last four games to be \(0.60,0.50,0.40\), and \(0.70\), respectively. (a) What are the chances that you will get to watch your team play on Jan. 1 ? (b) Is the probability that your team wins all four games given that they have won at least three games equal to the probability that they win the fourth game? Explain. (c) Is the probability that your team wins all four games given that they won the first three games equal to the probability that they win the fourth game?

An urn contains one white chip and a second chip that is equally likely to be white or black. A chip is drawn at random and returned to the urn. Then a second chip is drawn. What is the probability that a white appears on the second draw given that a white appeared on the first draw? (Hint: Let \(W_{i}\) be the event that a white chip is selected on the \(i\) th draw, \(i=1,2\). Then \(P\left(W_{2} \mid W_{1}\right)=\frac{P\left(W_{1} \cap W_{2}\right)}{P\left(W_{1}\right)}\). If both chips in the urn are white, \(P\left(W_{1}\right)=1\); otherwise, \(\left.P\left(W_{1}\right)=\frac{1}{2} .\right)\)

Imagine that the test tube pictured contains \(2 n\) grains of sand, \(n\) white and \(n\) black. Suppose the tube is vigorously shaken. What is the probability that the two colors of sand will completely separate; that is, all of one color fall to the bottom, and all of the other color lie on top? (Hint: Consider the \(2 n\) grains to be aligned in a row. In how many ways can the \(n\) white and the \(n\) black grains be permuted?)

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