/*! 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} Q66E Consider the following informati... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Q66E (page 85)

Consider the following information about travellers on vacation (based partly on a recent Travelocity poll): \({\rm{40\% }}\) check work email, \({\rm{30\% }}\) use a cell phone to stay connected to work, \({\rm{25\% }}\) bring a laptop with them, \({\rm{23\% }}\) both check work email and use a cell phone to stay connected, and \({\rm{51\% }}\) neither check work email nor use a cell phone to stay connected nor bring a laptop. In addition, \({\rm{88}}\) out of every \({\rm{100}}\) who bring a laptop also check work email, and \({\rm{70}}\) out of every \({\rm{100}}\) who use a cell phone to stay connected also bring a laptop. a. What is the probability that a randomly selected traveller who checks work email also uses a cell phone to stay connected? b. What is the probability that someone who brings a laptop on vacation also uses a cell phone to stay connected? c. If the randomly selected traveller checked work email and brought a laptop, what is the probability that he/ she uses a cell phone to stay connected?

Short Answer

Expert verified

a. \({\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}\mid {{\rm{A}}_{\rm{E}}}} \right){\rm{ = 0}}{\rm{.575}}\),

b. \({\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}\mid {{\rm{A}}_{\rm{L}}}} \right){\rm{ = 0}}{\rm{.84}}\),

c. \({\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}\mid {{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ = 0}}{\rm{.91}}\)

Step by step solution

01

Determine denote events and given probabilities

Denote events

\(\begin{array}{l}{{\rm{A}}_{\rm{E}}}{\rm{ = \{ check work mail \} }}\\{{\rm{A}}_{\rm{C}}}{\rm{ = \{ use cell phone to stay connected to work \} ;}}\\{{\rm{A}}_{\rm{L}}}{\rm{ = \{ bring a laptop \} }}\end{array}\)

we are given the following probabilities

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}} \right)\rm &=& 0{\rm{.4}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}} \right) \rm &=& 0{\rm{.3}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{L}}}} \right) \rm &=& 0{\rm{.25}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{C}}}} \right) \rm &=& 0{\rm{.23}}\\{\rm{P}}\left( {{\rm{A}}_{\rm{E}}^{\rm{垄}}{\rm{脟}A}_{\rm{C}}^{\rm{垄}}{\rm{脟}A}_{\rm{L}}^{\rm{垄}}} \right)\rm &=& 0{\rm{.51}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}\mid {{\rm{A}}_{\rm{L}}}} \right)\rm &=& 0{\rm{.88}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{L}}}\mid {{\rm{A}}_{\rm{C}}}} \right)\rm &=& 0{\rm{.7}}\end{array}\)

02

Determine the value of \({\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}\mid {{\rm{A}}_{\rm{E}}}} \right)\)

(a):

We need to find the probability that a randomly selected traveler who checks work email also uses a cell phone to stay connected, which is a conditional probability of \({{\rm{A}}_{\rm{C}}}\)given that the event \({{\rm{A}}_{\rm{E}}}\)has occurred. Therefore,

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}\mid {{\rm{A}}_{\rm{E}}}} \right)\rm &=& \frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}{\rm{脟}}{{\rm{A}}_{\rm{E}}}} \right)}}{{{\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}} \right)}}\\ \rm &=& \frac{{{\rm{0}}{\rm{.23}}}}{{{\rm{0}}{\rm{.4}}}}\\ \rm &=& 0{\rm{.575}}\end{array}\)

where we used the definition of conditional probability.

Conditional probability of \({\rm{A}}\) given that the event \({\rm{B}}\) has occurred, for which \({\rm{P}}\left( {\rm{B}} \right){\rm{ > 0,}}\)is

\({\rm{P(A}}\mid {\rm{B) = }}\frac{{{\rm{P(A{脟}B)}}}}{{{\rm{P(B)}}}}\)

for any two events \({\rm{A}}\)and\({\rm{B}}\).

03

Determine the value of \({\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}\mid {{\rm{A}}_{\rm{L}}}} \right)\)

(b):

The probability that someone who brings a laptop on vacation also uses a cell phone to stay connected is conditional probability of \({{\rm{A}}_{\rm{C}}}\)given that the event \({{\rm{A}}_{\rm{L}}}\)has occurred. We have

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}\mid {{\rm{A}}_{\rm{L}}}} \right)\mathop {\rm{ = }}\limits^{{\rm{(1)}}} \frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{L}}}{\rm{脟}}{{\rm{A}}_{\rm{C}}}} \right)}}{{{\rm{P}}\left( {{{\rm{A}}_{\rm{L}}}} \right)}}\\\mathop {\rm{ = }}\limits^{{\rm{(2)}}} \frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{L}}}\mid {{\rm{A}}_{\rm{C}}}} \right){\rm{*P}}\left( {{{\rm{A}}_{\rm{C}}}} \right)}}{{{\rm{P}}\left( {{{\rm{A}}_{\rm{L}}}} \right)}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.7*0}}{\rm{.3}}}}{{{\rm{0}}{\rm{.25}}}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.21}}}}{{{\rm{0}}{\rm{.25}}}}\\{\rm{ = 0}}{\rm{.84}}\end{array}\)

(1): here we use the definition of conditional probability, (2): here we use the multiplication rule given below.

The Multiplication Rule

\({\rm{P(A{脟}B) = P(A}}\mid {\rm{B)*P(B)}}\)

04

Determine the value of \({\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{É}}{{\rm{A}}_{\rm{C}}}{\rm{É}}{{\rm{A}}_{\rm{L}}}} \right)\)

(c):

The probability we are asked to calculate is conditional probability of \({{\rm{A}}_{\rm{C}}}\) given that the event \({{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}\) occurred. We have the following

\({\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}\mid {{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ = }}\frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}{\rm{脟}}{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right)}}{{{\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right)}}\)

Using the proposition given below, the following is true

\({\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脡}}{{\rm{A}}_{\rm{C}}}{\rm{脡}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{\rm{E}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{C}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{L}}}} \right){\rm{ - P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{C}}}} \right){\rm{ - P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ - P}}\left( {{{\rm{A}}_{\rm{C}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{C}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right)\)

or equally

\({\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{C}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脡}}{{\rm{A}}_{\rm{C}}}{\rm{脡}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ - P}}\left( {{{\rm{A}}_{\rm{E}}}} \right){\rm{ - P}}\left( {{{\rm{A}}_{\rm{C}}}} \right){\rm{ - P}}\left( {{{\rm{A}}_{\rm{L}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{C}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{C}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right)\)

We are missing

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脡}}{{\rm{A}}_{\rm{C}}}{\rm{脡}}{{\rm{A}}_{\rm{L}}}} \right)\rm &=& 1 - P\left( {{{\left( {{{\rm{A}}_{\rm{E}}}{\rm{脡}}{{\rm{A}}_{\rm{C}}}{\rm{脡}}{{\rm{A}}_{\rm{L}}}} \right)}^{\rm{垄}}}} \right)\\ \rm &=& 1 - P\left( {{{\rm{A}}_{\rm{C}}}{\rm{脟}}{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right)\\ \rm &=& 1 - 0{\rm{.51}}\\ \rm &=& 0{\rm{.49}}\end{array}\)

05

Determine the value of \({\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}\mid {{\rm{A}}_{\rm{E}}}{\rm{Ç}}{{\rm{A}}_{\rm{L}}}} \right)\)

Now we can substitute the values

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{C}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脡}}{{\rm{A}}_{\rm{C}}}{\rm{脡}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ - P}}\left( {{{\rm{A}}_{\rm{E}}}} \right){\rm{ - P}}\left( {{{\rm{A}}_{\rm{C}}}} \right){\rm{ - P}}\left( {{{\rm{A}}_{\rm{L}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{C}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{C}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{0}}{\rm{.49 - 0}}{\rm{.4 - 0}}{\rm{.3 - 0}}{\rm{.25 + 0}}{\rm{.22 + 0}}{\rm{.23 + 0}}{\rm{.21}}\\{\rm{ = 0}}{\rm{.2}}{\rm{.}}\end{array}\)

(1): here we calculated \({\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{l}}}} \right)\)as

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{\rm{L}}}} \right){\rm{*P}}\left( {{{\rm{A}}_{\rm{E}}}\mid {{\rm{A}}_{\rm{L}}}} \right)\\{\rm{ = 0}}{\rm{.25*0}}{\rm{.88}}\\{\rm{ = 0}}{\rm{.22,}}\end{array}\)

where the other probabilities were calculated previously.

Proposition: For every three events \({\rm{A,B}}\)and \({\rm{C}}\)the following is true

\({\rm{P(A脡 B脡 C) = P(A) + P(B) + P(C) - P(A脟B) - P(A脟C) - P(B脟C) + P(A脟B脟C)}}\)

Finally, we have

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}\mid {{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ = }}\frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}{\rm{脟}}{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right)}}{{{\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right)}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.2}}}}{{{\rm{0}}{\rm{.22}}}}\\{\rm{ = 0}}{\rm{.91}}\end{array}\)

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 91影视!

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

For customers purchasing a refrigerator at a certain appliance store, let A be the event that the refrigerator was manufactured in the U.S., B be the event that the refrigerator had an icemaker, and C be the event that the customer purchased an extended warranty. Relevant probabilities are

\(\begin{aligned}{\rm{P(A) = }}{\rm{.75}}\;\;\;{\rm{P(B}}\mid {\rm{A) = }}{\rm{.9}}\;\;\;{\rm{P}}\left( {{\rm{B}}\mid {{\rm{A}}^{\rm{\cent}}}} \right){\rm{ = }}{\rm{.8}}\\{\rm{P(C}}\mid {\rm{A\c{C}B) = }}{\rm{.8}}\;\;\;{\rm{P}}\left( {{\rm{C}}\mid {\rm{A\c{C}}}{{\rm{B}}^{\rm{\cent}}}} \right){\rm{ = }}{\rm{.6}}\\{\rm{P}}\left( {{\rm{C}}\mid {{\rm{A}}^{\rm{\cent}}}{\rm{\c{C}B}}} \right){\rm{ = }}{\rm{.7}}\;\;\;{\rm{P}}\left( {{\rm{C}}\mid {{\rm{A}}^{\rm{\cent}}}{\rm{\c{C}}}{{\rm{B}}^{\rm{\cent}}}} \right){\rm{ = }}{\rm{.3}}\end{aligned}\)

Construct a tree diagram consisting of first-, second-, and third-generation branches, and place an event label and appropriate probability next to each branch. b. Compute \({\rm{P(A\c{C}B\c{C}C)}}\). c. Compute \({\rm{P(B\c{C}C)}}\). d. Compute \({\rm{P(C)}}\). e. Compute \({\rm{P(A}}\mid {\rm{B\c{C}C)}}\), the probability of a U.S. purchase given that an icemaker and extended warranty are also purchased

A starting lineup in basketball consists of two guards, two forwards, and a center.

a. A certain college team has on its roster three centers, four guards, four forwards, and one individual (X) who can play either guard or forward. How many different starting lineups can be created? (Hint: Consider lineups without X, then lineups with X as guard, then lineups with X as forward.)

b. Now suppose the roster has \({\rm{5}}\) guards, \({\rm{5}}\) forwards, \({\rm{3}}\)centers, and \({\rm{2}}\) 鈥渟wing players鈥� (X and Y) who can play either guard or forward. If 5 of the \({\rm{15}}\) players are randomly selected, what is the probability that they constitute a legitimate starting lineup?

In five-card poker, a straight consists of five cards with adjacent denominations (e.g., \({\rm{9}}\)of clubs, \({\rm{10}}\)of hearts, jack of hearts, queen of spades, and king of clubs). Assuming that aces can be high or low, if you are dealt a five-card hand, what is the probability that it will be a straight with high card \({\rm{10}}\)? What is the probability that it will be a straight? What is the probability that it will be a straight flush (all cards in the same suit)?

The three most popular options on a certain type of newcar are a built-in GPS (A), a sunroof (B), and an automatictransmission (C). If 40% of all purchasers request A, 55% request B, 70% request C, 63% request Aor B,77% request Aor C, 80% request Bor C, and 85% request Aor Bor C, determine the probabilities of the following events. (Hint:鈥淎or B鈥� is the event that at leastone of the two options is requested; try drawing a Venn

diagram and labeling all regions.)

a. The next purchaser will request at least one of thethree options.

b. The next purchaser will select none of the three options.

c. The next purchaser will request only an automatictransmission and not either of the other two options.

d. The next purchaser will select exactly one of thesethree options.

The Reviews editor for a certain scientific journal decides whether the review for any particular book should be short (1鈥�2 pages), medium (3鈥�4 pages), or long (5鈥�6 pages). Data on recent reviews indicates that \({\rm{60\% }}\) of them are short, \({\rm{30\% }}\)are medium, and the other \({\rm{10\% }}\)are long. Reviews are submitted in either Word or LaTeX. For short reviews, \({\rm{80\% }}\)are in Word, whereas \({\rm{50\% }}\)of medium reviews are in Word and \({\rm{30\% }}\) of long reviews are in Word. Suppose a recent review is randomly selected. a. What is the probability that the selected review was submitted in Word format? b. If the selected review was submitted in Word format, what are the posterior probabilities of it being short, medium, or long?
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation
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.