/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 Let \(S=\\{a, b, c, d, e, f\\}\)... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(S=\\{a, b, c, d, e, f\\}\) be a sample space of an experiment and let \(E=\\{a, b\\}, F=\\{a, d, f\\}\), and \(G=\) \(\\{b, c, e\\}\) be events of this experiment. Find the events \(E^{c}\) and \(F^{c} \cap G\).

Short Answer

Expert verified
In conclusion, the required events are: \(E^{c} = \{c, d, e, f\}\) \(F^{c} \cap G = \{b, c, e\}\).

Step by step solution

01

Understanding Complements

Complements are the elements in the sample space that are not included in the given event. To find the complement of an event, we will identify which elements of the sample space are not in the event.
02

Find the Complement of Event E

Given that \(E=\{a, b\}\), we need to find the elements in the sample space that are not in E. We can see that \(\{c, d, e, f\}\) are the remaining elements in the sample space not included in E, and therefore \(E^{c}=\{c, d, e, f\}\).
03

Find the Complement of Event F

Similarly, to find the complement of event F, we look for elements in the sample space not in the set F. Given that \(F=\{a, d, f\}\), we can see that the elements not included in F are \(\{b, c, e\}\). So, \(F^{c}=\{b, c, e\}\).
04

Intersection of \(F^{c}\) and G

Now, we need to find the intersection of \(F^{c}\) and G. The intersection of two sets is the set of elements that are common to both sets. \(F^{c}=\{b, c, e\}\) and \(G=\{b, c, e\}\). The elements that are common to both sets are \(\{b, c, e\}\). Therefore, the intersection of \(F^{c}\) and G is \(F^{c} \cap G = \{b, c, e\}\). In conclusion, the required events are: \(E^{c} = \{c, d, e, f\}\) \(F^{c} \cap G = \{b, c, e\}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Sample Space
In probability theory, a sample space is a fundamental concept that represents the set of all possible outcomes of a particular experiment. It's like the universe of everything that could happen. For instance, if you were rolling a six-sided die, your sample space would include all numbers from 1 to 6. In our example, the sample space is denoted by \[S = \{a, b, c, d, e, f\}.\]This implies that anything that could happen in this experiment must be within this set. Accurately identifying a sample space is critical since it lays the foundation for determining probabilities of various events that could occur based on the experiment.
Event Complements
An event in probability is any subset of the sample space. When working with probability, often we discuss the idea of event complements. The complement of an event is simply what the event does not include within the sample space. For example, if event \( E \) includes outcomes \( \{a, b\} \), its complement \( E^c \) includes everything in the sample space not in \( E \). So, for \( E^c \): - Start with the sample space \( \{a, b, c, d, e, f\} \)- Remove elements of \( E \)- Results in \( \{c, d, e, f\} \)This means \( E^c \) encompasses all scenarios where the event \( E \) does not occur. Understanding event complements is crucial because sometimes it's simpler to calculate the probability of an event not happening and subtract from 1 to find the probability of it happening.
Set Intersection
Set intersections in probability help us understand shared elements between two events. The intersection of two sets, say \( A \) and \( B \), is the set containing all elements that are common to both \( A \) and \( B \). It is symbolized as \( A \cap B \).In our scenario, we have the sets:- \( F^c = \{b, c, e\} \)- \( G = \{b, c, e\} \)Finding the intersection involves identifying which elements appear in both. Clearly,\[F^c \cap G = \{b, c, e\}.\]Intersections are useful in probability to determine simultaneous occurrences of two events, aiding in complex probability calculations.
Elementary Event Operations
Elementary event operations are the building blocks of probability problems and involve operations like finding complements, intersections, and unions of events. Each operation gives insight into different types of questions you might ask about events in a probability experiment.
  • Complements: Determine what does not happen when a particular event occurs.
  • Intersections: Check what occurs simultaneously in multiple events.
  • Unions: Find at least one event occurrence, focusing on collective effects.
In our example, the operations performed ( - Complement of \( E \): \( E^c = \{c, d, e, f\} \)
- Intersection of \( F^c \)and \( G \): \( F^c \cap G = \{b, c, e\}\)) demonstrate elementary event operations. These basic operations are fundamental for more advanced probability concepts like calculating probabilities for independent or dependent events using set operations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Based on data obtained from the National Institute of Dental Research, it has been determined that \(42 \%\) of 12 -yr-olds have never had a cavity, \(34 \%\) of 13 -yr-olds have never had a cavity, and \(28 \%\) of 14 -yrolds have never had a cavity. Suppose a child is selected at random from a group of 24 junior high school students that includes six 12 -yr-olds, eight 13 -yr-olds, and ten 14 -yrolds. If this child does not have a cavity, what is the probability that this child is 14 yrs old?

CAR THEFT Figures obtained from a city's police department seem to indicate that, of all motor vehicles reported as stolen, \(64 \%\) were stolen by professionals whereas \(36 \%\) were stolen by amateurs (primarily for joy rides). Of those vehicles presumed stolen by professionals, \(24 \%\) were recovered within \(48 \mathrm{hr}, 16 \%\) were recovered after \(48 \mathrm{hr}\), and \(60 \%\) were never recovered. Of those vehicles presumed stolen by amateurs, \(38 \%\) were recovered within \(48 \mathrm{hr}, 58 \%\) were recovered after \(48 \mathrm{hr}\), and \(4 \%\) were never recovered. a. Draw a tree diagram representing these data. b. What is the probability that a vehicle stolen by a professional in this city will be recovered within \(48 \mathrm{hr}\) ? c. What is the probability that a vehicle stolen in this city will never be recovered?

Quaurr CoNrroL. An automobile manufacturer obtains the microprocessors used to regulate fuel consumption in its automobiles from three microelectronic firms: \(\mathrm{A}, \mathrm{B}\), and C. The quality-control department of the company has determined that \(1 \%\) of the microprocessors produced by firm \(A\) are defective, \(2 \%\) of those produced by firm \(B\) are defective, and \(1.5 \%\) of those produced by firm \(\mathrm{C}\) are defective. Firms \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) supply \(45 \%, 25 \%\), and \(30 \%\), respectively, of the microprocessors used by the company. What is the probability that a randomly selected automobile manufactured by the company will have a defective microprocessor?

A nationwide survey conducted by the National Cancer Society revealed the following information. Of 10,000 people surveyed, 3200 were "heavy coffee drinkers" and 160 had cancer of the pancreas. Of those who had cancer of the pancreas, 132 were heavy coffee drinkers. Using the data in this survey, determine whether the events "being a heavy coffee drinker" and "having cancer of the pancreas" are independent events.

The probabilitics that the three patients who are scheduled to receive kidney transplants at General Hospital will suffer rejection are \(\frac{1}{2}, \frac{1}{3}\) and \(\frac{1}{10}\). Assuming that the cvents (kidney rejection) are indcpendent, find the probability that a. At least one paticnt will suffer rejection. b. Exactly two patients will suffer rejection.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.