/*! 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} Q19E Human visual inspection of solde... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Q19E (page 65)

Human visual inspection of solder joints on printed circuit boards can be very subjective. Part of the problem stems from the numerous types of solder defects (e.g., pad non wetting, knee visibility, voids) and even the degree to which a joint possesses one or more of these defects. Consequently, even highly trained inspectors can disagree on the disposition of a particular joint. In one batch of 10,000 joints, inspector A found 724 that were judged defective, inspector B found 751 such joints, and 1159 of the joints were judged defective by at least one of the inspectors. Suppose that one of the 10,000 joints is randomly selected.

a. What is the probability that the selected joint was judged to be defective by neither of the two inspectors?

b. What is the probability that the selected joint was judged to be defective by inspector B but not by inspector A?

Short Answer

Expert verified

a. The probability that the selected joint was judged to be defective by neither of the two inspectors is 0.8841.

b. The probability that the selected joint was judged to be defective by inspector B but not by inspector A is 0.0435.

Step by step solution

01

Given information

The number of joints randomly selected is 10000.

The number of defective joints found by inspector A is 724.

The number of defective joints found by inspector B is 751.

The number of defective joints that were found by at least one of the inspectors is 1159.

02

Compute the probability

Let A be the event that the number of defective joints is found by inspector A.

Let B be the event that the number of defective joints is found by inspector B.

The event \(A \cup B\) represents the event of number of defective joints that were found by at least one of the inspectors.

The probability of event A is computed as,

\(\begin{aligned}P\left( A \right) &= \frac{{724}}{{10000}}\\ &= 0.0724\end{aligned}\)

Therefore, the probability of event A is 0.0724.

The probability of event B is computed as,

\(\begin{aligned}P\left( B \right) &= \frac{{751}}{{10000}}\\ &= 0.0751\end{aligned}\)

Therefore, the probability of event B is 0.0751.

The probability of event \(A \cup B\) is computed as,

\(\begin{aligned}P\left( {A \cup B} \right) &= \frac{{1159}}{{10000}}\\ &= 0.1159\end{aligned}\)

Therefore, the probability of event \(A \cup B\) is 0.1159.

Using the formula for the probability of union of the two events,

\(P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\)

The probability that the number of defective joints that were found by inspector A and inspector B is computed as,

\(\begin{aligned}P\left( {A \cap B} \right) &= P\left( A \right) + P\left( B \right) - P\left( {A \cup B} \right)\\ &= 0.0724 + 0.0751 - 0.1159\\ &= 0.0316\end{aligned}\)

Therefore, the probability that the number of defective joints that were found by inspector A and inspector B is 0.0316

a.

The probability that the selected joint was judged to be defective by neither of the two inspectors is computed as,

\(\begin{aligned}P\left( {A \cup B} \right)' &= 1 - P\left( {A \cup B} \right)\\ &= 1 - 0.1159\\ &= 0.8841\end{aligned}\)

Therefore, the probability that the selected joint was judged to be defective by neither of the two inspectors is 0.8841.

b.

The probability that the selected joint was judged to be defective by inspector B but not by inspector A is computed as,

\(\begin{aligned}P\left( {B \cup A'} \right) &= P\left( B \right) - P\left( {A \cap B} \right)\\ &= 0.0751 - 0.0316\\ &= 0.0435\end{aligned}\)

Therefore, the probability that the selected joint was judged to be defective by inspector B but not by inspector A is 0.0435

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

An academic department has just completed voting by secret ballot for a department head. The ballot box contains four slips with votes for candidate Aand three slips with votes for candidate B.Suppose these slips are removed from the box one by one.

a. List all possible outcomes.

b. Suppose a running tally is kept as slips are removed. For what outcomes does Aremain ahead of B throughout the tally?

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?

An insurance company offers four different deductible levels鈥攏one, low, medium, and high鈥攆or its home owner鈥檚 policy holders and three different levels鈥攍ow, medium, and high鈥攆or its automobile policyholders. The accompanying table gives proportions for the various categories of policyholders who have both types of insurance. For example, the proportion of individuals with both low homeowner鈥檚 deductible and low auto deductible is .06(6% of all such individuals).

贬辞尘别辞飞苍别谤鈥檚

Auto N L M H

L .04 .06 .05 .03

M .07 .10 .20 .10

H .02 .03 .15 .15

Suppose an individual having both types of policies is randomly selected.

a. What is the probability that the individual has a medium auto deductible and a high homeowner鈥檚 deductible?

b. What is the probability that the individual has a low auto deductible? A low homeowner鈥檚 deductible?

c. What is the probability that the individual is in the same category for both auto and homeowner鈥檚 deductibles?

d. Based on your answer in part (c), what is the probability that the two categories are different?

e. What is the probability that the individual has at least one low deductible level?

f. Using the answer in part (e), what is the probability that neither deductible level is low?

Consider randomly selecting a student at a large university, and let Abe the event that the selected student has a Visa card and Bbe the analogous event for MasterCard. Suppose that P(A)=.6 and P(B)=.4.

a. Could it be the case that P(A\( \cap \)B)=.5? Why or why not? (Hint:See Exercise 24.)

b. From now on, suppose that P(A\( \cap \)B)=.3. What is the probability that the selected student has at least one of these two types of cards?

c. What is the probability that the selected student has neither type of card?

d. Describe, in terms of Aand B, the event that the selected student has a Visa card but not a MasterCard, and then calculate the probability of this event.

e. Calculate the probability that the selected student has exactly one of the two types of cards.

A system consists of two identical pumps, \(\# 1\) and \(\# 2\). If one pump fails, the system will still operate. However, because of the added strain, the remaining pump is now more likely to fail than was originally the case. That is, \(r\; = P\left( {\# 2\;fails|\# 1 fails} \right) > P\left( {\# 2 fails} \right) = q\). If at least one pump fails by the end of the pump design life in \(7\% \) of all systems and both pumps fail during that period in only 1%, what is the probability that pump \(\# 1\)will fail during the pump design life?
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Decision Maths

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.