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Eine M眉nze wird zweimal hintereinander geworfen. Dabei kann jeweils Kopf oder Zahl geworfen werden. (a) Aus wie viel Elementen besteht die von allen m枚glichen Elementarereignissen erzeugte \(\sigma\)-Ereignisalgebra \(S_{0} ?\) (b) Aus welchen Ereignissen besteht die von den Ereignissen \(A=\), Der erste Wurf ist Kopf" und \(B=\), Es wurde mindes tens einmal Kopf geworfen" erzeugte \(\sigma\)-Ereignisalgebra \(S_{1}\) ? Enth?lt \(S_{1}\) auch: \(C=\),Der zweite Wurf ist Kopf"?

Short Answer

Expert verified
Answer: There are 11 elements in the event-algebra 饾憜0. The event-algebra 饾憜1 does not contain event C.

Step by step solution

01

List all possible elementary events

Before we can analyze the 蟽-补濒驳别产谤补, we need to list all possible elementary events. The coin can land heads (H) or tails (T). Since the coin is flipped twice, we can have the following combinations of outcomes: (H, H), (H, T), (T, H), and (T, T).
02

Find the size of the 蟽-补濒驳别产谤补 generated by all possible elementary events 饾憜0

Now, we need to find the size of the event-algebra generated by all possible elementary events. The 蟽-补濒驳别产谤补 contains the empty set, the universal set, and all possible combinations of the elementary events. In this case, the 蟽-补濒驳别产谤补 饾憜0 contains: 1. The empty set: 鈭 2. The universal set: {(H, H), (H, T), (T, H), (T, T)} 3. All possible combinations of elementary events: - {(H, H)} - {(H, T)} - {(T, H)} - {(T, T)} - {(H, H), (H, T)} - {(H, H), (T, H)} - {(H, H), (T, T)} - {(T, H), (H, T)} - {(T, H), (T, T)} - {(H, T), (T, T)} - {(H, H), (H, T), (T, H)} So, we have a total of 11 elements in the event-algebra 饾憜0.
03

Define events A, B, and C

Now, we need to define the three events mentioned in the second part of the problem: A: "The first flip is heads" 鈫 {(H, H), (H, T)} B: "At least one flip is heads" 鈫 {(H, H), (H, T), (T, H)} C: "The second flip is heads" 鈫 {(H, H), (T, H)}
04

Find the 蟽-补濒驳别产谤补 generated by events A and B: 饾憜1

We will now find the combination of events that generate the 蟽-补濒驳别产谤补 饾憜1: 1. The empty set: 鈭 2. The set of all elementary events: {(H, H), (H, T), (T, H), (T, T)} 3. The individual sets: - {(H, H), (H, T)} 鈫 Event A - {(H, H), (H, T), (T, H)} 鈫 Event B 4. The intersection of A and B: - {(H, H), (H, T)} 鈭 {(H, H), (H, T), (T, H)} = {(H, H), (H, T)} So, 饾憜1 contains 4 elements: 鈭, {(H, H), (H, T), (T, H), (T, T)}, {(H, H), (H, T)}, and {(H, H), (H, T), (T, H)}.
05

Check if 饾憜1 contains event C

To check if 饾憜1 contains event C=("The second flip is heads"), we simply look at the elements in 饾憜1. We see that event C = {(H, H), (T, H)} is not in 饾憜1. Therefore, 饾憜1 doesn't contain event C.

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Key Concepts

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

Understanding Probability Theory
Probability theory is the branch of mathematics that deals with quantifying the likelihood of different outcomes in uncertain events. It forms the foundation for statistical inference and helps us make predictions based on incomplete information. Key to this field is the use of mathematical structures called 蟽-补濒驳别产谤补蝉, which provide a rigorous way to define and manipulate sets of probabilities.

At its core, probability is measured on a scale from 0 to 1, where 0 means an event will never occur, and 1 indicates certainty. In between these two extremes lays a spectrum of probabilities, reflecting the range of potential outcomes for any given situation. In the case of flipping a coin, where only two outcomes are possible鈥攈eads or tails鈥攅ach outcome has an equal probability of occurring, which is 0.5 when assuming a fair coin. However, probability theory also equips us to handle more complex scenarios involving multiple events and outcomes.
Elementary Events and Their Role
An elementary event in probability theory is one of the basic outcomes that can't be broken down into simpler components. For the coin flip example, elementary events include landing on heads (H) or tails (T) in one flip. When considering multiple sequential flips, the elementary events become combinations of these initial possibilities, like (H, H) or (T, H) for two successive flips.

Understanding and listing all possible elementary events is crucial when we want to analyze all that could happen in an experiment. They form the building blocks for more complex events, termed 'event combinations,' and the entirety of these elementary events make up the sample space of a probability problem. By exploring these fundamental outcomes, we can construct the universe of all possible scenarios before delving into how they may interact or combine to form the full landscape of an experiment's outcomes.
Event Combinations and 蟽-Algebra Construction
In probability theory, event combinations refer to the various ways in which elementary events can be grouped or combined to form compound events. These include unions, intersections, and complements of elementary events. For example, the event 'at least one flip is heads' combines several elementary events into a larger set. Event combinations are significant because they encompass all the possible outcomes that can occur together in an experiment.

A 蟽-补濒驳别产谤补 is a collection of such event combinations that satisfy certain mathematical properties like closure under complementation and countable unions. It essentially helps to organize the possible outcomes in a way that's conducive to applying the laws of probability. The construction of a 蟽-补濒驳别产谤补 includes simple events and all their possible combinations including the empty set and the sample space. This organized structure ensures that probabilities can be assigned consistently and logically, setting the stage for more advanced probability theory concepts and their applications to real-world situations.

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Most popular questions from this chapter

(a) Auf wie viel verschiedene Arten lassen sich \(m\) verschiedene Kugeln auf \(n\) verschiedene Schubladen aufteilen? (b) Auf wie viel verschiedene Arten lassen sich \(m\) gleiche Kugeln auf \(n\) verschiedene Schubladen aufteilen?

1\. An der Frankfurter B枚rse wurde eine Gruppe von 70 Wertpapierbesitzern befragt. Es stellte sich heraus, dass 50 von ihnen Aktien und 40 Pfandbriefe besitzen. Wie viele der Befragten besitzen sowohl Aktien als auch Pfandbriefe? 2\. Aus einer zweiten Umfrage unter allen Rechtsanw盲lten in Frankfurt wurde bekannt, dass \(60 \%\) der Anw盲lte ein Haus und \(80 \%\) ein Auto besitzen. \(20 \%\) der Anw盲lte sind Mitglied einer Partei. Von allen Befragten sind \(40 \%\) Auto- und Hausbesitzer, \(10 \%\) Autobesitzer und Mitglied einer Partei und \(15 \%\) Hausbesitzer und Mitglied einer Partei. Wie viel Prozent besitzen sowohl eine Auto als auch ein Haus und sind Mitglied einer Partei?

Bei einem M眉nz-Wurf-Spiel wird eine M眉nze hintereinander mehrmals geworfen, die mit Wahrscheinlichkeit \(\gamma\) 鈥濳opf" wirft. Dabei seien die W眉rfe total unabh盲ngig voneinander. Wird ,,Kopf" geworfen, erhalten Sie einen Euro, wird .,Zahl" geworfen, zahlen Sie einen Euro. Sie starten mit \(0 \in\). Das Spiel bricht ab, wenn Ihr Spielkonto entweder ein Guthaben von \(2 \in\) oder Schulden von \(2 \in\) aufweist. Wie gro \(B\) ist die Wahrscheinlichkeit \(\alpha\), dass Sie mit einem Guthaben von \(2 拢\) das Spiel beenden?

Wir betrachten vier Spielkarten \(B\) ? Bube, \(D \triangleq\) Dame, \(K\) \triangleq K枚nig und den Joker \(\triangleq J .\) Jede dieser vier Karten werde mit gleicher Wahrscheinlichkeit \(\frac{1}{4}\) gezogen. Der Joker kann als Bube, Dame oder K枚nig gewertet werden. Wir ziehen. eine Karte und definieren die drei Ereignisse: $$ \begin{aligned} &b=\\{B \cup J\\} \quad \Rightarrow \quad \mathrm{P}(b)=\frac{1}{2} \\ &d=\\{D \cup J\\} \quad \Rightarrow \quad \mathrm{P}(d)=\frac{1}{2} \\ &k=\\{K \cup J\\} \quad \Rightarrow \quad \mathrm{P}(k)=\frac{1}{2} \end{aligned} $$ Zeigen Sie: Die Ereignisse \(b, d, k\) sind paarweise, aber nicht total unabh?ngig.

Zeige: Sind \(A\) und \(B\) unabh盲ngig, dann sind auch \(A\) und \(B^{C}\) unabh?ngig, ebenso \(B\) und \(A^{\text {C }}, A^{C}\) und \(B^{C}\)

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