/*! 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 92 Consider the following informati... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider the following information about travelers on vacation: \(40 \%\) check work e-mail, \(30 \%\) use a cell phone to stay connected to work, \(25 \%\) bring a laptop with them on vacation, \(23 \%\) both check work e-mail and use a cell phone to stay connected, and \(51 \%\) neither check work e-mail nor use a cell phone to stay connected nor bring a laptop. In addition \(88 \%\) of those who bring a laptop also check work e-mail and \(70 \%\) of those who use a cell phone to stay connected also bring a laptop. With \(E=\) event that a traveler on vacation checks work e-mail, \(C=\) event that a traveler on vacation uses a cell phone to stay connected, and \(L=\) event that a traveler on vacation brought a laptop, use the given information to determine the following probabilities. A Venn diagram may help. a. \(P(E)\) b. \(P(C)\) c. \(P(L)\) d. \(P(E\) and \(C)\) e. \(P\left(E^{C}\right.\) and \(C^{C}\) and \(\left.L^{C}\right)\) f. \(P(E\) and \(\operatorname{Cor} L)\) g. \(\quad P(E \mid L)\) h. \(P(L \mid C)\) i. \(P(E\) and \(C\) and \(L)\) j. \(P(E\) and \(L)\) k. \(P(C\) and \(L)\) 1\. \(P(C \mid E\) and \(L)\)

Short Answer

Expert verified
a. 0.4, b. 0.3, c. 0.25, d. 0.23, e. 0.51, f. 0.16, g. 0.88, h. 0.7, i. 0.16, j. 0.22, k. 0.21, l. 0.73.

Step by step solution

01

Understanding the Problem

Firstly, define the events accurately: E = A vacationer checks work email, C = A vacationer uses a cell phone, L = A vacationer brings a laptop. The probabilities of these single events are given directly in the problem. Similarly, we can identify the intersection events EC, EL and CL. Note that the problem statement also gives value for the event where none E, C or L occurs.
02

Calculating Single Event Probabilities

Given directly in the problem, we use these as our base: \n a. \(P(E) = 0.4\), \n b. \(P(C) = 0.3\), \n c. \(P(L) = 0.25\)
03

Calculating Intersection Probabilities

Given in the problem statement: \(P(E \cap C) = 0.23\). Probabilities \(P(E \cap L)\) and \(P(C \cap L)\) are not directly given, we will derive them in the next steps.
04

Calculating \(P(E \cap L)\) and \(P(C \cap L)\)

Given that 88% of those who bring a laptop also check work email, \(P(E \cap L) = 0.88 * P(L) \) = 0.22. \n And, 70% of those who use a cell phone also bring a laptop, \(P(C \cap L) = 0.70 * P(C) \) = 0.21.
05

Calculating Union of All three events

By the principle of inclusion and exclusion: \(P(E \cup C \cup L) = P(E) + P(C) + P(L) - P(E \cap C) - P(E \cap L) - P(C \cap L) + P(E \cap C \cap L)\). We can substitute the known values and solve for \(P(E \cap C \cap L)\) which turns out to be 0.16.
06

Calculating Probability of Complement Event

Given that 51% neither check work e-mail, nor use a cell phone nor bring a laptop, e. \(P(E^C \cap C^C \cap L^C) = 0.51\).
07

Calculating Conditional Probabilities

From the definition of Conditional Probability we know that \(P(A | B) = P(A \cap B) / P(B)\), this can be used to find: \n g. \(P(E | L) = P(E \cap L) / P(L) = 0.88\), \n h. \(P(L | C) = P(L \cap C) / P(C) = 0.7\), \n l. \(P(C | E \cap L) = P(E \cap C \cap L) / P(E \cap L) = 0.73\)
08

Result

The solution to probabilities are as follows: \n a. \(P(E) = 0.4\), \n b. \(P(C) = 0.3\), \n c. \(P(L) = 0.25\), \n d. \(P(E \cap C)= 0.23\), \n e. \(P(E^C \cap C^C \cap L^C) = 0.51\), \n f. \(P(E \cap C \cap L) = 0.16\), \n g. \(P(E | L) = 0.88\), h. \(P(L | C) = 0.7\), \n i. \(P(E \cap C \cap L) = 0.16\), \n j. \(P(E \cap L) = 0.22\), \n k. \(P(C \cap L) = 0.21\), \n l. \(P(C | E \cap L) = 0.73\)

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.

Conditional Probability
Conditional probability is a measure of the likelihood that an event will occur given that another event has already happened. When we say 'the probability of A given B,' we are finding out how probable event A is when event B is known to be true. The formal definition uses the probability of both events happening together divided by the probability of the event that is known to have occurred, that is, \[\begin{equation} P(A | B) = \frac{P(A \cap B)}{P(B)},\end{equation}\] if the probability of B is not zero. Conditional probability can dramatically change the likelihood of an event compared to its unconditional or 'simple' probability because it includes additional information related to the outcome of the event.
In the context of the textbook exercise, the calculation of conditional probability comes into play when determining the likelihood of certain events concerning others. For instance, the problem calculates the conditional probability of travelers checking work email given they brought a laptop, expressed as \[\begin{equation} P(E | L).\end{equation}\] This takes into account only those travelers with laptops and isolates the probability within this subgroup. Such calculations are critical in fields where outcomes are interdependent, such healthcare, finance, and social sciences.
Venn Diagram
A Venn diagram is an illustration that uses circles to show the relationships among different sets. Each circle represents a set, and the overlap between circles represents elements that are common to the sets that overlap. Venn diagrams are exceptionally useful in probability theory because they help to visualize complex relationships and intersections between events.
In our example with travelers, you can create a Venn diagram with three circles representing the events E (checking work email), C (using a cell phone), and L (bringing a laptop). The overlaps show the intersection of these events. For instance, the area where circles E and C overlap represents vacationers who both check work emails and use a cellphone, which corresponds to \[\begin{equation} P(E \cap C).\end{equation}\] A Venn diagram is an essential visualization tool to comprehend the various probability questions posed in the exercise. Not only does it aid in understanding the given probabilities, but it also allows students to reason about the problems visually. This can be particularly helpful when dealing with the complexities of probability theory.
Inclusion-Exclusion Principle
The inclusion-exclusion principle is a fundamental concept that provides a way of counting elements in combined sets by adding the sizes of individual sets and then subtracting the sizes of the intersecting parts to avoid overcounting. Applied to probability, the inclusion-exclusion principle allows us to calculate the probability of the union of multiple events by ensuring that common elements between events are not counted multiple times.
For calculating the probability of any one of the events E, C, or L happening, the principle is stated as follows:\[\begin{equation} P(E \cup C \cup L) = P(E) + P(C) + P(L) - P(E \cap C) - P(E \cap L) - P(C \cap L) + P(E \cap C \cap L).\end{equation}\]This equation represents the union of all three events, taking into account the necessary additions and subtractions. Utilizing the inclusion-exclusion principle in the exercise helps determine \[\begin{equation} P(E \cap C \cap L),\end{equation}\] the probability that all three events occur together. Understanding this principle is vital for solving more complex probability problems, where multiple events with intersections need to be considered, making it a cornerstone concept in combinatorics and probability theory.

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

The article "Birth Beats Long Odds for Leap Year Mom, Baby" (San Luis Obispo Tribune, March 2 , 1996) reported that a leap year baby (someone born on February 29) became a leap year mom when she gave birth to a baby on February 29,1996 . The article stated that a hospital spokesperson said that only about 1 in \(2.1\) million births is a leap year baby born to a leap year mom (a probability of approximately .00000047). a. In computing the given probability, the hospital spokesperson used the fact that a leap day occurs only once in 1461 days. Write a few sentences explaining how the hospital spokesperson computed the stated probability. b. To compute the stated probability, the hospital spokesperson had to assume that the birth was equally likely to occur on any of the 1461 days in a four- year period. Do you think that this is a reasonable assumption? Explain. c. Based on your answer to Part (b), do you think that the probability given by the hospital spokesperson is too small, about right, or too large? Explain.

Approximately \(30 \%\) of the calls to an airline reservation phone line result in a reservation being made. a. Suppose that an operator handles 10 calls. What is the probability that none of the 10 calls result in a reservation? b. What assumption did you make to calculate the probability in Part (a)? c. What is the probability that at least one call results in a reservation being made?

The Australian newspaper The Mercury (May 30 , 1995) reported that, based on a survey of 600 reformed and current smokers, \(11.3 \%\) of those who had attempted to quit smoking in the previous 2 years had used a nicotine aid (such as a nicotine patch). It also reported that \(62 \%\) of those who quit smoking without a nicotine aid began smoking again within 2 weeks and \(60 \%\) of those who used a nicotine aid began smoking again within 2 weeks. If a smoker who is trying to quit smoking is selected at random, are the events selected smoker who is trying to quit uses a nicotine aid and selected smoker who bas attempted to quit begins smoking again within 2 weeks independent or dependent events? Justify your answer using the given information.

A theater complex is currently showing four \(\mathrm{R}\) rated movies, three \(\mathrm{PG}-13\) movies, two \(\mathrm{PG}\) movies, and one G movie. The following table gives the number of people at the first showing of each movie on a certain Saturday: $$ \begin{array}{ccc} \text { Theater } & \text { Rating } & \begin{array}{c} \text { Number of } \\ \text { Viewers } \end{array} \\ \hline 1 & \mathrm{R} & 600 \\ 2 & \mathrm{PG}-13 & 420 \\ 3 & \mathrm{PG}-13 & 323 \\ 4 & \mathrm{R} & 196 \\ 5 & \mathrm{G} & 254 \\ 6 & \mathrm{PG} & 179 \\ 7 & \mathrm{PG}-13 & 114 \\ 8 & \mathrm{R} & 205 \\ 9 & \mathrm{R} & 139 \\ 10 & \mathrm{PG} & 87 \\ \hline \end{array} $$ Suppose that a single one of these viewers is randomly selected. a. What is the probability that the selected individual saw a PG movie? b. What is the probability that the selected individual saw a \(\mathrm{PG}\) or a \(\mathrm{PG}-13\) movie? c. What is the probability that the selected individual did not see an \(\mathrm{R}\) movie?

A new model of laptop computer can be ordered with one of three screen sizes ( 10 inches, 12 inches, 15 inches) and one of four hard drive sizes \((50 \mathrm{~GB}, 100 \mathrm{~GB}\), \(150 \mathrm{~GB}\), and \(200 \mathrm{~GB}\) ). Consider the chance experiment in which a laptop order is selected and the screen size and hard drive size are recorded. a. Display possible outcomes using a tree diagram. b. Let \(A\) be the event that the order is for a laptop with a screen size of 12 inches or smaller. Let \(B\) be the event that the order is for a laptop with a hard drive size of at most \(100 \mathrm{~GB}\). What outcomes are in \(A^{C}\) ? In \(A \cup B ?\) In \(A \cap B\) ? c. Let \(C\) denote the event that the order is for a laptop with a \(200 \mathrm{~GB}\) hard drive. Are \(A\) and \(C\) disjoint events? Are \(B\) and \(C\) disjoint?

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.