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Chapter 2: Problem 77

An aircraft seam requires 25 rivets. The seam will have to be reworked if any of these rivets is defective. Suppose rivets are defective independently of one another, each with the same probability. a. If \(20 \%\) of all seams need reworking, what is the probability that a rivet is defective? b. How small should the probability of a defective rivet be to ensure that only \(10 \%\) of all seams need reworking?

Short Answer

Expert verified
a. 0.00936 b. 0.00412

Step by step solution

01

Understanding the Problem

We want to find the probability that a rivet is defective given that the probability a seam needs reworking is 20%. The seam needs reworking if at least one of the 25 rivets is defective.
02

Establishing the Probability of No Defective Rivets

Let the probability of a rivet being defective be \( p \). The probability that a single rivet is not defective is \( 1-p \). Hence, the probability that none of the 25 rivets are defective is \( (1-p)^{25} \).
03

Using Complement Rule for Reworked Seams

The probability that at least one rivet is defective (seam needs reworking) is the complement of having no defective rivets: \( 1 - (1-p)^{25} \). We are given this probability is 0.2.
04

Setting Up Equation to Find Defective Probability

Equation: \[ 1 - (1 - p)^{25} = 0.2 \]. Simplifying, we have: \[ (1 - p)^{25} = 0.8 \].
05

Solving for \( p \)

Take the 25th root of both sides: \( 1-p = 0.8^{1/25} \). Calculate this using a calculator to find \( p \): \( p = 1 - 0.8^{1/25} \).
06

Calculating Defective Probability for Part a

After solving, we find \( p \approx 0.00936 \). This is the probability of a defective rivet for a 20% rework rate.
07

Reworking for 10% Seam Issue

Repeat steps by setting \( 1 - (1-p)^{25} = 0.1 \), leading to \( (1-p)^{25} = 0.9 \).
08

Calculating Required Defective Probability for Part b

Take the 25th root: \( 1-p = 0.9^{1/25} \). Solving, we find \( p \approx 0.00412 \). This is the necessary probability for a rivet to be defective to ensure only 10% rework rate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Defective Probability
In probability theory, 'defective probability' refers to the likelihood that an individual item, such as a rivet, is defective. In our context, it is critical because a seam of an aircraft will need reworking if any of its 25 rivets is defective. This depends on the probability that a single rivet fails.
Understanding defective probability is crucial in practical applications where quality control is necessary. If the probability is high, more thorough inspections and quality improvements are needed. Conversely, a lower defective probability means better quality and reliability of the components, such as rivets in this aviation example.
Complement Rule
The complement rule is an essential concept in probability theory. It helps you determine the probability of an event by understanding its opposite. If you know the probability that an event does not happen, you can easily find the probability that it does happen by subtracting this from 1.
In the scenario with rivets, if we denote the probability that no rivet is defective as \((1-p)^{25}\), then the probability that at least one rivet is defective, which is the reason for reworking a seam, is \(1 - (1-p)^{25}\). This application shows how using complements can simplify calculations effectively.
It's vital in many real-world applications where calculating the direct probability of a complex event can be daunting.
Equation Solving
Equation solving in probability problems often involves setting up equations based on given conditions and using algebraic techniques to find unknown values. In our exercise, we aim to find out the probability \(p\) that a rivet is defective.
By using known probabilities (0.2 or 0.1 for reworking a seam), we form equations like \((1-p)^{25} = 0.8\) or \((1-p)^{25} = 0.9\). Solving these equations involves isolating \(p\).
The process frequently includes using calculators to compute expressions like \(0.8^{1/25}\), yielding the defective probability \(p\). This problem-solving skill is not only applicable to probability but to any field requiring analytical reasoning.
Probability Calculation
Probability calculation involves determining how likely an event is to occur. This calculation often requires both understanding the situation and applying mathematical principles.
In this scenario, calculating the probability requires knowledge of how multiple independent events combine, which is modeled using powers like \((1-p)^{25}\). This technique allows for determining complex probabilities based on simpler single events.
When you compute the defective probability, first find the probability that no rivet is defective, then use the complement rule to assess the likelihood of at least one being defective. Such calculations are foundational in fields like risk assessment, decision-making, and quality control.

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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 $$ \begin{aligned} &P(A)=.75 \quad P(B \mid A)=.9 \quad P\left(B \mid A^{\prime}\right)=.8 \\ &P(C \mid A \cap B)=.8 \quad P\left(C \mid A \cap B^{\prime}\right)=.6 \\ &P\left(C \mid A^{\prime} \cap B\right)=.7 \quad P\left(C \mid A^{\prime} \cap B^{\prime}\right)=.3 \end{aligned} $$ a. 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 \(P(A \cap B \cap C)\). c. Compute \(P(B \cap C)\). d. Compute \(P(C)\). e. Compute \(P(A \mid B \cap C)\), the probability of a U.S. purchase given that an icemaker and extended warranty are also purchased.

A production facility employs 20 workers on the day shift, 15 workers on the swing shift, and 10 workers on the graveyard shift. A quality control consultant is to select 6 of these workers for in-depth interviews. Suppose the selection is made in such a way that any particular group of 6 workers has the same chance of being selected as does any other group (drawing 6 slips without replacement from among 45). a. How many selections result in all 6 workers coming from the day shift? What is the probability that all 6 selected workers will be from the day shift? b. What is the probability that all 6 selected workers will be from the same shift? c. What is the probability that at least two different shifts will be represented among the selected workers? d. What is the probability that at least one of the shifts will be unrepresented in the sample of workers?

A certain factory operates three different shifts. Over the last year, 200 accidents have occurred at the factory. Some of these can be attributed at least in part to unsafe working conditions, whereas the others are unrelated to working conditions. The accompanying table gives the percentage of accidents falling in each type of accidentshift category. $$ \begin{array}{llcc} & & \begin{array}{c} \text { Unsafe } \\ \text { Conditions } \end{array} & \begin{array}{c} \text { Unrelated } \\ \text { to Conditions } \end{array} \\ \hline \text { Shift } & \text { Day } & 10 \% & 35 \% \\ & \text { Swing } & 8 \% & 20 \% \\ \text { Night } & 5 \% & 22 \% \\ \hline \end{array} $$ Suppose one of the 200 accident reports is randomly selected from a file of reports, and the shift and type of accident are determined. a. What are the simple events? b. What is the probability that the selected accident was attributed to unsafe conditions? c. What is the probability that the selected accident did not occur on the day shift?

An ATM personal identification number (PIN) consists of four digits, each a \(0,1,2, \ldots 8\), or 9 , in succession. a. How many different possible PINs are there if there are no restrictions on the choice of digits? b. According to a representative at the author's local branch of Chase Bank, there are in fact restrictions on the choice of digits. The following choices are prohibited: (i) all four digits identical (ii) sequences of consecutive ascending or descending digits, such as 6543 (iii) any sequence starting with 19 (birth years are too easy to guess). So if one of the PINs in (a) is randomly selected, what is the probability that it will be a legitimate PIN (that is, not be one of the prohibited sequences)? c. Someone has stolen an ATM card and knows that the first and last digits of the PIN are 8 and 1, respectively. He has three tries before the card is retained by the ATM (but does not realize that). So he randomly selects the \(2^{\text {nd }}\) and \(3^{\text {rd }}\) digits for the first try, then randomly selects a different pair of digits for the second try, and yet another randomly selected pair of digits for the third try (the individual knows about the restrictions described in (b) so selects only from the legitimate possibilities). What is the probability that the individual gains access to the account? d. Recalculate the probability in (c) if the first and last digits are 1 and 1 , respectively.

Fifteen telephones have just been received at an authorized service center. Five of these telephones are cellular, five are cordless, and the other five are corded phones. Suppose that these components are randomly allocated the numbers \(1,2, \ldots, 15\) to establish the order in which they will be serviced. a. What is the probability that all the cordless phones are among the first ten to be serviced? b. What is the probability that after servicing ten of these phones, phones of only two of the three types remain to be serviced? c. What is the probability that two phones of each type are among the first six serviced?
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