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Chapter 1: Problem 25

Die 4 Seiten eines Tetraeders seien wie folgt gef盲rbt: Fl?che I rot, Fl盲che II blau, Fl盲che III gr眉n, Fl盲che IV rot, blau und gr眉n gleichzeitig. Der Tetraeder werde geworfen. Man pr眉fe, ob die Ereignisse, die unten liegende Fl盲che enth盲lt die rote, blaue bzw. gr眉ne Farbe paarweise bzw. vollst盲ndig (stoch.) unab. h盲ngig sind.

Short Answer

Expert verified
The events are pairwise independent but not mutually independent.

Step by step solution

01

- Identify Sample Space

The sample space consists of the four faces of the tetrahedron. Let's denote these faces by I, II, III, IV.
02

- Define Events

Define the events as follows: - Event A: The bottom face is red (I or IV). - Event B: The bottom face is blue (II or IV). - Event C: The bottom face is green (III or IV).
03

- Calculate Probabilities

Calculate the probabilities of each event: - P(A): There are 2 red faces out of 4, so P(A) = 2/4 = 1/2. - P(B): There are 2 blue faces out of 4, so P(B) = 2/4 = 1/2. - P(C): There are 2 green faces out of 4, so P(C) = 2/4 = 1/2.
04

- Calculate Joint Probabilities

Calculate the probabilities of the intersections of each pair of events: - P(A and B): Only face IV is both red and blue. P(A and B) = 1/4. - P(A and C): Only face IV is both red and green. P(A and C) = 1/4. - P(B and C): Only face IV is both blue and green. P(B and C) = 1/4.
05

- Verify Pairwise Independence

Check if each pair of events is independent by confirming that P(A and B) = P(A)P(B), P(A and C) = P(A)P(C), and P(B and C) = P(B)P(C): - P(A)P(B) = (1/2)*(1/2) = 1/4. Since P(A and B) = 1/4, A and B are independent. - P(A)P(C) = (1/2)*(1/2) = 1/4. Since P(A and C) = 1/4, A and C are independent. - P(B)P(C) = (1/2)*(1/2) = 1/4. Since P(B and C) = 1/4, B and C are independent.
06

- Verify Mutual Independence

Check if the events are mutually independent by confirming that P(A and B and C) = P(A)P(B)P(C): - P(A and B and C) = Probability that face IV is on the bottom, which is 1/4. - P(A)P(B)P(C) = (1/2)*(1/2)*(1/2) = 1/8. Since 1/4 鈮� 1/8, the events are not mutually independent.

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Key Concepts

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

Sample Space
In probability theory, the sample space is the set of all possible outcomes of a random experiment. For this problem, the experiment is rolling a tetrahedron with four distinct faces. The tetrahedron has faces I, II, III, and IV. Each face can land on the bottom when the tetrahedron is rolled. The sample space for this experiment is: \[ S = \text{{\textlbrace I, II, III, IV \textbraceright}} \] Each outcome in the sample space is equally likely, providing a foundational base for calculating event probabilities.
Event Probability
An event is a subset of the sample space that includes one or more possible outcomes. Event probability refers to the likelihood of an event occurring. Let's consider the following events:
  • Event A: The bottom face is red (I or IV).
  • Event B: The bottom face is blue (II or IV).
  • Event C: The bottom face is green (III or IV).
To find the probability of each event, count the favorable outcomes and divide by the total number of outcomes. For example, there are 2 faces with red, so \[ P(A) = \frac{2}{4} = \frac{1}{2} \] Do the same for events B and C, and you'll find \[ P(B) = \frac{1}{2} \] and \[ P(C) = \frac{1}{2} \]
Intersection Probability
Intersection probability deals with the likelihood of two or more events occurring simultaneously. The intersection of events A and B, denoted by \(A \text{ and } B\), occurs when the bottom face is both red and blue. For our tetrahedron, only face IV is both red and blue. Therefore, \[ P(A \text{ and } B) = \frac{1}{4} \] When calculating intersection probabilities for pairs of events in this problem, it is evident that only one face satisfies both conditions for each pair:
  • \( P(A \text{ and } C) = \frac{1}{4} \)
  • \( P(B \text{ and } C) = \frac{1}{4} \)
Pairwise Independence
Two events are pairwise independent if the occurrence of one does not affect the probability of the other. Formally, events A and B are independent if \[ P(A \text{ and } B) = P(A) P(B) \] For our tetrahedron:
  • \( P(A) P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)
  • \( P(A \text{ and } B) = \frac{1}{4} \)
Thus, events A and B are independent. Similarly, you can confirm independence for \(A\) and \(C\), and \(B\) and \(C\).
Mutual Independence
Mutual independence involves more than just pairs of events. Events A, B, and C are mutually independent if every combination of events also independently holds true. This means: \[ P(A \text{ and } B \text{ and } C) = P(A) P(B) P(C) \] For our tetrahedron: \[ P(A) P(B) P(C) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \] However, \[ P(A \text{ and } B \text{ and } C) = \frac{1}{4} \] Since \frac{1}{4} otequals \frac{1}{8}\, events A, B, and C are not mutually independent.

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