/*! 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 5 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. Are the events \(E\) and \(F\) mutually exclusive?

Short Answer

Expert verified
No, the events E and F are not mutually exclusive, since their intersection is not empty: Intersection(E, F) = {a}.

Step by step solution

01

Identify the events E and F

Event E = {a, b} Event F = {a, d, f}
02

Check for the intersection of events E and F

We will now determine the intersection of E and F. The intersection of two sets is the set of elements that are common to both sets. Intersection(E, F) = {a}
03

Determine if the events E and F are mutually exclusive

Since the intersection of events E and F contains an element, a, the events E and F are not mutually exclusive.

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, a **sample space** is a fundamental concept. It's like the foundation of any probability problem. The sample space is the set of all possible outcomes of an experiment. In our exercise example, we are given the sample space \( S = \{a, b, c, d, e, f\} \).

This means that when we conduct the experiment, every result we get will be one of these elements in the set. It's crucial to identify this at the beginning of any problem because all probability calculations will be based on these possibilities.

  • Ensure all possible outcomes are considered.
  • Each event like \(E\), \(F\), and \(G\) is a subset of this sample space.
  • Think of the sample space as a map for the event outcomes to exist within.
Intersection of Sets
Understanding the **intersection of sets** is key when working with probabilities. The intersection refers to the elements common to both sets. In mathematical terms, if \(E\) and \(F\) are subsets of our sample space, then the intersection \(E \cap F\) is the set of elements they share.

For instance, in our exercise:
  • Event \(E = \{a, b\}\)
  • Event \(F = \{a, d, f\}\)
When we find \(E \cap F\), we look for common elements in both sets \(E\) and \(F\). The result here is \(\{a\}\), indicating that "a" is the only outcome common to both events. This concept helps us determine the nature of the relationship between the two events.

Understanding intersections is vital for distinguishing whether events can happen simultaneously or not.
Probability Concepts
At the heart of this problem is the idea of **probability concepts**, which involves determining how likely events are to occur. Here, we focus on "mutually exclusive events."

Mutually exclusive events are those that cannot happen at the same time. In simpler terms, no outcome can belong to both events. If their intersection has no elements (is an empty set), we call them mutually exclusive. For events \(E\) and \(F\), when we calculated the intersection: \(E \cap F = \{a\}\), we observed that it was not empty because the element "a" was common in both events.

  • This indicates \(E\) and \(F\) are not mutually exclusive.
  • The presence of any element in the intersection negates mutual exclusivity.
This understanding is crucial in probability as it influences how we calculate combined probabilities of events.

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

A time study was conducted by the production manager of Vista Vision to determine the length of time in minutes required by an assembly worker to complete a certain task during the assembly of its Pulsar color television sets. a. Describe a sample space corresponding to this time study. b. Describe the event \(E\) that an assembly worker took 2 min or less to complete the task. c. Describe the event \(F\) that an assembly worker took more than 2 min to complete the task.

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. If \(A\) is a subset of \(B\), then \(P(A) \leq P(B)\).

In a poll conducted among 2000 college freshmen to ascertain the political views of college students. the accompanying data were obtained. Determine the empirical probability distribution associated with these data. $$ \begin{array}{lccccc} \hline \text { Political Views } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } \\ \hline \text { Respondents } & 52 & 398 & 1140 & 386 & 24 \\ \hline \end{array} $$ A: Far left B: Liberal C: Middle of the road D: Conservative E: Far right

A study conducted by the Corrections Department of a certain state revealed that 163,605 people out of a total adult population of \(1,778,314\) were under correctional supervision (on probation, on parole, or in jail). What is the probability that a person selected at random from the adult population in that state is under correctional supervision?

According to a study of 100 drivers in metropolitan Washington, D.C., whose cars were equipped with cameras with sensors, the distractions and the number of incidents (crashes, near crashes, and situations that require an evasive maneuver after the driver was distracted) caused by these distractions are as follows: $$ \begin{array}{lccccccccc} \hline \text { Distraction } & A & B & C & D & E & F & G & H & I \\ \hline \text { Driving Incidents } & 668 & 378 & 194 & 163 & 133 & 134 & 111 & 111 & 89 \\ \hline \end{array} $$ where \(A=\) Wireless device (cell phone, PDA) \(\begin{aligned} B &=\text { Passenger } \\ C &=\text { Something inside car } \\ D &=\text { Vehicle } \\ E &=\text { Personal hygiene } \\ F &=\text { Eating } \\ G &=\text { Something outside car } \\ H &=\text { Talking/singing } \\ I &=\text { Other } \end{aligned}\) If an incident caused by a distraction is picked at random, what is the probability that it was caused by a. The use of a wireless device? b. Something other than personal hygiene or eating?

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.