/*! 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 145 Terry \& Sons makes bearings... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 145

Terry \& Sons makes bearings for autos. The production system involves two independent processing machines so that each bearing passes through these two processes. The probability that the first processing machine is not working properly at any time is \(.08\), and the probability that the second machine is not working properly at any time is \(.06\). Find the probability that both machines will not be working properly at any given time.

Short Answer

Expert verified
The probability that both machines will not be working properly at any given time is 0.0048.

Step by step solution

01

Identify independent events and their probabilities

The text tells us that the two processing machines are independent. This means that the performance of one machine does not affect the performance of the other. The given probabilities are: \n\n1. The probability that the first machine is not working properly, denoted \(P(M1)\), is \(0.08\).\n2. The probability that the second machine is not working properly, denoted \(P(M2)\), is \(0.06\).
02

Apply the rule of multiplication for independent events

The rule of multiplication for independent events states that the probability of two independent events A and B occurring together is given by the product of their individual probabilities. So, we need to calculate the product \(P(M1) * P(M2)\).
03

Calculate the desired probability

Substituting the provided values into our formula, we obtain: \n\n\( P(M1 \text{ and } M2) = 0.08 * 0.06 = 0.0048 \).

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

Refer to Exercise 4.48. A 2010-2011 poll conducted by Gallup (www.gallup.com/poll/148994/ Emotional-Health-Higher-Among-Older- Americans.aspx) examined the emotional health of a large number of Americans. Among other things, Gallup reported on whether people had Emotional Health Index scores of 90 or higher, which would classify them as being emotionally well-off. The report was based on a survey of 65,528 people in the age group \(35-44\) years and 91,802 people in the age group \(65-74\) years. The following table gives the results of the survey, converting percentages to frequencies. $$ \begin{array}{lcc} \hline & \text { Emotionally Well-Off } & \text { Emotionally Not Well-Off } \\\ \hline \text { 35-44 Age group } & 16,016 & 49,512 \\ \text { 65-74 Age group } & 32,583 & 59,219 \\ \hline \end{array} $$ a. Suppose that one person is selected at random from this sample of 157,330 Americans. Find the following probabilities. i. \(P(35-44\) age group and emotionally not well-off \()\) ii. \(P(\) emotionally well-off and \(65-74\) age group \()\) b. Find the joint probability of the events \(35-44\) age group and \(65-74\) age group. Is this probability zero? Explain why or why not.

Two thousand randomly selected adults were asked if they think they are financially better off than their parents. The following table gives the two- way classification of the responses based on the education levels of the persons included in the survey and whether they are financially better off, the same as. or worse off than their parents. $$ \begin{array}{lccc} \hline & \begin{array}{c} \text { Less Than } \\ \text { High School } \end{array} & \begin{array}{c} \text { High } \\ \text { School } \end{array} & \begin{array}{c} \text { More Than } \\ \text { High School } \end{array} \\ \hline \text { Better off } & 140 & 450 & 420 \\ \text { Same as } & 60 & 250 & 110 \\ \text { Worse off } & 200 & 300 & 70 \\ \hline \end{array} $$ Suppose one adult is selected at random from these 2000 adults. Find the following probabilities. a. \(P\) (better off or high school) b. \(P\) (more than high school or worse off) c. \(P(\) better off or worse off \()\)

What is the joint probability of two mutually exclusive events? Give one example.

There is an area of free (but illegal) parking near an inner-city sports arena. The probability that a car parked in this area will be ticketed by police is \(.35\), that the car will be vandalized is \(.15\), and that it will be ticketed and vandalized is \(.10 .\) Find the probability that a car parked in this area will be ticketed or vandalized.

Two thousand randomly selected adults were asked whether or not they have ever shopped on the Internet. The following table gives a two-way classification of the responses. $$ \begin{array}{lcc} \hline & \text { Have Shopped } & \text { Have Never Shopped } \\ \hline \text { Male } & 500 & 700 \\ \text { Female } & 300 & 500 \\ \hline \end{array} $$ Suppose one adult is selected at random from these 2000 adults. Find the following probabilities. a. \(P(\) has never shopped on the Internet or is a female) b. \(P(\) is a male \(o r\) has shopped on the Internet) c. \(P\) (has shopped on the Internet or has never shopped on the Internet)
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

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.