Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 2: Q64E (page 84)

The Reviews editor for a certain scientific journal decides whether the review for any particular book should be short (1–2 pages), medium (3–4 pages), or long (5–6 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?

Short Answer

Expert verified

Respectively the posterior probabilities of short, medium, long is

\(\begin{aligned}P\left( {{A_1}\mid W} \right) &= 0.727;\\P\left( {{A_2}\mid W} \right) &= 0.227;\\P\left( {{A_3}\mid W} \right) &= 0.046\end{aligned}\)

Step by step solution

01

Determine the first and second layer of event

The Multiplication Rule

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

First layer of event is

\(\begin{aligned}{{\rm{A}}_{\rm{1}}}\rm &= \{ short book \} \\{{\rm{A}}_{\rm{2}}}\rm &= \{ medium book \} \\{{\rm{A}}_{\rm{3}}}\rm &= \{ long book \} \end{aligned}\)

and the second layer events are

\(\begin{aligned}\rm W &= \{ submited in word \} \\\rm L &= \{ submited in LaTeX \} \end{aligned}\)

The adequate probabilities for the first layer are

\(\begin{aligned}{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ = 0}}{\rm{.6}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.3}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{3}}}} \right){\rm{ = 0}}{\rm{.1}}\end{aligned}\)

02

Determine multiplication rule to calculate every of the four intersections displayed in the tree diagram

The adequate conditional probabilities are

\(\begin{aligned}{\rm{P}}\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{1}}}} \right)\rm &= 0{\rm{.8}}\\{\rm{P}}\left( {{\rm{L}}\mid {{\rm{A}}_{\rm{1}}}} \right)\rm &= 1 - P\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{1}}}} \right)\rm = 0{\rm{.2}}\\{\rm{P}}\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{2}}}} \right)\rm &= 0{\rm{.5}}\\{\rm{P}}\left( {{\rm{L}}\mid {{\rm{A}}_{\rm{2}}}} \right)\rm &= 1 - P\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{2}}}} \right)\rm = 0{\rm{.5}}\\{\rm{P}}\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{3}}}} \right)\rm &= 0{\rm{.3}}\\{\rm{P}}\left( {{\rm{L}}\mid {{\rm{A}}_{\rm{3}}}} \right)\rm &= 1 - P\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{3}}}} \right)\rm = 0{\rm{.7}}\end{aligned}\)

03

Determine the probability of intersection

The tree diagram also contains the probabilities of the intersections. We use the multiplication rule to calculate every of the four intersections displayed in the tree diagram

\(\begin{aligned}{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{{C}W}}} \right)\rm &= P\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{*P}}\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{1}}}} \right)\rm = 0{\rm{.6*0}}\rm.8 = 0{\rm{.48}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{{C}L}}} \right)\rm &= P\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{*P}}\left( {{\rm{L}}\mid {{\rm{A}}_{\rm{1}}}} \right)\rm = 0{\rm{.6*0}}\rm.2 = 0{\rm{.12}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}{\rm{{C}W}}} \right)\rm &= P\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{*P}}\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{2}}}} \right)\rm = 0{\rm{.3*0}}\rm.5 = 0{\rm{.15}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}{\rm{{C}L}}} \right)\rm &= P\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{*P}}\left( {{\rm{L}}\mid {{\rm{A}}_{\rm{2}}}} \right)\rm = 0{\rm{.3*0}}\rm.5 = 0{\rm{.15}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{3}}}{\rm{{C}W}}} \right)\rm &= P\left( {{{\rm{A}}_{\rm{3}}}} \right){\rm{*P}}\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{3}}}} \right)\rm = 0{\rm{.1*0}}\rm.3 = 0{\rm{.03}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{3}}}{\rm{{C}L}}} \right)\rm &= P\left( {{{\rm{A}}_{\rm{3}}}} \right){\rm{*P}}\left( {{\rm{L}}\mid {{\rm{A}}_{\rm{3}}}} \right)\rm = 0{\rm{.1*0}}\rm.7 = 0{\rm{.07}}\end{aligned}\)

From the tree diagram given below, we will calculate everything easier.

04

Determine the law of total probability

(a):

We need to find the probability that the selected review was submitted in Word format

\({\rm{P(W)}}\)The Law of Total Probability

If \({{\rm{A}}_{\rm{1}}}{\rm{,}}{{\rm{A}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{A}}_{\rm{n}}}\) are mutually exclusive, and \({\rm{E }}_{{\rm{i = 1}}}^{\rm{n}}{{\rm{A}}_{\rm{i}}}{\rm{ = \Omega }}\), then for any event \({\rm{B}}\) the following is true\(\begin{aligned}\rm P(B) &= P\left( {{\rm{B}}\mid {{\rm{A}}_{\rm{1}}}} \right){\rm{P}}\left({{{\rm{A}}_{\rm{1}}}} \right){\rm{ + P}}\left( {{\rm{B}}\mid {{\rm{A}}_{\rm{2}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ + \ldots + P}}\left( {{\rm{B}}\mid {{\rm{A}}_{\rm{n}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{n}}}} \right)\\\rm &=\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\rm{P}} \left( {{\rm{B}}\mid {{\rm{A}}_{\rm{i}}}} \right){\rm{P}}\left({{{\rm{A}}_{\rm{i}}}} \right)\end{aligned}\)

For \({\rm{B = W}}\) we have

\(\begin{aligned}\rm P(W) &= P\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{1}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ + P}}\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{2}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ + P}}\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{3}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{3}}}} \right)\\\rm &= 0{\rm{.48 + 0}}{\rm{.15 + 0}}{\rm{.03}}\\\rm &= 0{\rm{.66}}\end{aligned}\)

05

Explain the Bayes’ theorem 

(b):

Bayes' Theorem

For \({{\rm{A}}_{\rm{1}}}{\rm{,}}{{\rm{A}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{A}}_{\rm{n}}}\) mutually exclusive events, and \({\rm{E }}_{{\rm{i = 1}}}^{\rm{n}}{{\rm{A}}_{\rm{i}}}{\rm{ = \Omega }}\) the prior probabilities are\(\left. {{\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}} \right){\rm{,i = 1,2, \ldots ,n}}} \right)\). If \({\rm{B}}\)is event for which\({\rm{P(B) > 0}}\), then the posterior probability of \({{\rm{A}}_{\rm{i}}}\)given that \({\rm{B}}\) has occurred

\({\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}\mid {\rm{B}}} \right){\rm{ = }}\frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}{\rm{{C}B}}} \right)}}{{{\rm{P(B)}}}}{\rm{ = }}\frac{{{\rm{P}}\left( {{\rm{B}}\mid {{\rm{A}}_{\rm{j}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{j}}}} \right)}}{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\rm{P}} \left( {{\rm{B}}\mid {{\rm{A}}_{\rm{i}}}} \right){\rm{*P}}\left( {{{\rm{A}}_{\rm{i}}}} \right)}}{\rm{,}}\;\;\;{\rm{j = 1,2, \ldots ,n}}{\rm{.}}\)

From the theorem, the posterior probabilities are

\(\begin{aligned}{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}\mid {\rm{W}}} \right)\rm &= \frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{{C}W}}} \right)}}{{{\rm{P(W)}}}}\\\rm &= \frac{{{\rm{0}}{\rm{.48}}}}{{{\rm{0}}{\rm{.66}}}}\\\rm &= 0{\rm{.727;}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}\mid {\rm{W}}} \right)\rm &= \frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}{\rm{{C}W}}} \right)}}{{{\rm{P(W)}}}}\\\rm &= \frac{{{\rm{0}}{\rm{.15}}}}{{{\rm{0}}{\rm{.66}}}}\\\rm &= 0{\rm{.227;}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{3}}}\mid {\rm{W}}} \right)\rm &= \frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{3}}}{\rm{{C}W}}} \right)}}{{{\rm{P(W)}}}}\\\rm &= \frac{{{\rm{0}}{\rm{.03}}}}{{{\rm{0}}{\rm{.66}}}}\\\rm &= 0{\rm{.046}}\end{aligned}\)

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Consider the system of components connected as in the accompanying picture. Components \({\rm{1}}\) and \({\rm{2}}\) are connected in parallel, so that subsystem works iff either 1 or 2 works; since \({\rm{3}}\)and\({\rm{4}}\) are connected in series, that subsystem works iff both\({\rm{3}}\)and\({\rm{4}}\)work. If components work independently of one another and P(component i works) \({\rm{ = }}{\rm{.9}}\)for \({\rm{i = }}{\rm{.1,2}}\)and \({\rm{ = }}{\rm{.8}}\)for \({\rm{i = 3,4}}\),calculate P(system works).

There has been a great deal of controversy over the last several years regarding what types of surveillance are appropriate to prevent terrorism. Suppose a particular surveillance system has a \({\rm{99\% }}\) chance of correctly identifying a future terrorist and a \({\rm{99}}{\rm{.9\% }}\)chance of correctly identifying someone who is not a future terrorist. If there are \({\rm{1000}}\) future terrorists in a population of \({\rm{300}}\) million, and one of these \({\rm{300}}\) million is randomly selected, scrutinized by the system, and identified as a future terrorist, what is the probability that he/she actually is a future terrorist? Does the value of this probability make you uneasy about using the surveillance system? Explain.

A large operator of timeshare complexes requires anyone interested in making a purchase to first visit the site of interest. Historical data indicates that \({\rm{20\% }}\) of all potential purchasers select a day visit, \({\rm{50\% }}\) choose a one-night visit, and \({\rm{30\% }}\)opt for a two-night visit. In addition, \({\rm{10\% }}\) of day visitors ultimately make a purchase, \({\rm{30\% }}\) of one-night visitors buy a unit, and \({\rm{20\% }}\) of those visiting for two nights decide to buy. Suppose a visitor is randomly selected and is found to have made a purchase. How likely is it that this person made a day visit? A one-night visit? A two-night visit?

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 \({\rm{2}}\) “swing players” (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?

A certain shop repairs both audio and video components. Let A denote the event that the next component brought in for repair is an audio component, and let B be the event that the next component is a compact disc player (so the event B is contained in A). Suppose that \(P\left( A \right) = 0.6\)and\(P\left( B \right) = 0.05\). What is\(P\left( {B|A} \right)\)?
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Probability and Statistics

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.

Sign-up for free