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Chapter 2: Problem 17

Hydraulic landing assemblies coming from an aircraft rework facility are each inspected for defects. Historical records indicate that \(8 \%\) have defects in shafts only, \(6 \%\) have defects in bushings only, and \(2 \%\) have defects in both shafts and bushings. One of the hydraulic assemblies is selected randomly. What is the probability that the assembly has a. a bushing defect? b. a shaft or bushing defect? c. exactly one of the two types of defects? d. neither type of defect?

Short Answer

Expert verified
a. 8% b. 16% c. 14% d. 84%

Step by step solution

01

Understand Given Data

We are given probabilities for defects in hydraulic assemblies: - Defects in shafts only: 8% - Defects in bushings only: 6% - Defects in both shafts and bushings: 2%. Define: \( P(S) \) = Probability of shaft defect, \( P(B) \) = Probability of bushing defect, and \( P(S \cap B) \) = Probability of both defects.
02

Calculate P(Bushing Defect)

The probability of a bushing defect can be found by adding the probabilities of having bushing defects only and having both defects: \[ P(B) = P(B \text{ only}) + P(S \cap B) = 0.06 + 0.02 = 0.08. \]
03

Calculate P(Shaft or Bushing Defect)

To find the probability of either a shaft or bushing defect, use the formula for the union of two events: \[ P(S \cup B) = P(S) + P(B) - P(S \cap B). \] Given that \( P(S) = 0.08 + 0.02 = 0.10 \) and \( P(B) = 0.08 \), calculate: \[ P(S \cup B) = 0.10 + 0.08 - 0.02 = 0.16. \]
04

Calculate P(Exactly One Type of Defect)

To find the probability of exactly one type of defect, sum the probabilities of a defect in shafts only and in bushings only. These are mutually exclusive events: \[ P(S \text{ only}) = 0.08 \] \[ P(B \text{ only}) = 0.06 \] Add them: \[ P(S \text{ only}) + P(B \text{ only}) = 0.08 + 0.06 = 0.14. \]
05

Calculate P(Neither Defect)

To find the probability of neither a shaft nor a bushing defect, subtract the probability of a defect from 1: \[ P( ext{neither}) = 1 - P(S \cup B) = 1 - 0.16 = 0.84. \]

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Key Concepts

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

Shaft Defect Probability
The concept of shaft defect probability deals with the likelihood that a hydraulic landing assembly has a defect in the shafts only. According to historical data, 8% of these assemblies have defects solely in the shafts. This means when we randomly select an assembly from a batch, there is an 8% chance it will be defective in the shaft.

To calculate this, use the given probability directly as provided in the exercise. Keep in mind that this probability is independent of the condition of the bushings in the assembly. Understanding this helps clarify that no matter what defects may be present in the bushings, the probability for shafts remains constant at 8%.
Bushing Defect Probability
Bushing defect probability focuses on finding out how likely it is that a hydraulic assembly has a defect in the bushings. From the exercise, we know that 6% of the assemblies have defects only in the bushings, and an additional 2% have both shaft and bushing defects.

To find the total probability of a bushing defect:
  • Consider defects only in bushings: 6%
  • Add those with both shaft and bushing defects: 2%
Thus, the probability (\( P(B) \)) of a bushing defect is 8% as calculated by: \[ P(B) = P(B \text{ only}) + P(S \cap B) = 0.06 + 0.02 = 0.08 \]
This helps in understanding that whenever we encounter a bushing defect, it may or may not be associated with a shaft defect.
Events Union Probability
The union probability of two events refers to the probability that either one event occurs, or both occur. In this context, it is about computing the chance that an assembly has a defect in either the shaft or the bushing, or both.

To calculate the union probability (\( P(S \cup B) \)) for shaft and bushing defects:
  • Use the probability of shaft defects: 10% (8% for only shaft and 2% shared with bushing)
  • Add the probability of bushing defects: 8%
  • Subtract the overlap (both defects): 2%
Apply the formula:\[ P(S \cup B) = P(S) + P(B) - P(S \cap B) \]Insert the known values:\[ P(S \cup B) = 0.10 + 0.08 - 0.02 = 0.16 \]
This helps us see that there is a 16% overall chance of encountering an assembly with either type of defect or both, which is a critical concept in probability theory.
Mutually Exclusive Events
Mutually exclusive events are those that cannot happen at the same time. In the case of hydraulic assembly defects, the events 'defect in shafts only' and 'defect in bushings only' are mutually exclusive.

To find the probability of one of these mutually exclusive events, i.e., exactly one of the two defects, simply add the probabilities of each event:
  • Defects in shafts only: 8%
  • Defects in bushings only: 6%
Thus, the probability of exactly one type of defect is:\[ P(S \text{ only}) + P(B \text{ only}) = 0.08 + 0.06 = 0.14 \]
This addition works because these events cannot occur together, making them completely independent of one another.

Grasping this concept allows us to handle scenarios in probability where outcomes are entirely separate from each other, simplifying complex problems significantly.

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Most popular questions from this chapter

An oil prospecting firm hits oil or gas on \(10 \%\) of its drillings. If the firm drills two wells, the four possible simple events and three of their associated probabilities are as given in the accompanying table. Find the probability that the company will hit oil or gas a. on the first drilling and miss on the second. b. on at least one of the two drillings.

A survey of consumers in a particular community showed that \(10 \%\) were dissatisfied with plumbing jobs done in their homes. Half the complaints dealt with plumber \(A\), who does \(40 \%\) of the plumbing jobs in the town. Find the probability that a consumer will obtain a. an unsatisfactory plumbing job, given that the plumber was \(A\). b. a satisfactory plumbing job, given that the plumber was \(A\).

If two events, \(A\) and \(B\), are such that \(P(A)=.5, P(B)=.3,\) and \(P(A \cap B)=.1\), find the following: a. \(P(A | B)\) b. \(P(B | A)\) c. \(P(A | A \cup B)\) d. \(P(A | A \cap B)\) e. \(P(A \cap B | A \cup B)\)

Americans can be quite suspicious, especially when it comes to government conspiracies. On the question of whether the U.S. Air Force has withheld proof of the existence of intelligent life on other planets, the proportions of Americans with varying opinions are given in the table. $$\begin{array}{lc} \hline \text { Opinion } & \text { Proportion } \\ \hline \text { Very likely } & .24 \\ \text { Somewhat likely } & .24 \\ \text { Unlikely } & .40 \\ \text { Other } & .12 \\ \hline \end{array}$$ Suppose that one American is selected and his or her opinion is recorded. a. What are the simple events for this experiment? b. Are the simple events that you gave in part (a) all equally likely? If not, what are the probabilities that should be assigned to each? c. What is the probability that the person selected finds it at least somewhat likely that the Air Force is withholding information about intelligent life on other planets?

A business office orders paper supplies from one of three vendors, \(V_{1}, V_{2},\) or \(V_{3} .\) Orders are to be placed on two successive days, one order per day. Thus, \(\left(V_{2}, V_{3}\right)\) might denote that vendor \(V_{2}\) gets the order on the first day and vendor \(V_{3}\) gets the order on the second day. a. List the sample points in this experiment of ordering paper on two successive days. b. Assume the vendors are selected at random each day and assign a probability to each sample point. c. Let \(A\) denote the event that the same vendor gets both orders and \(B\) the event that \(V_{2}\) gets at least one order. Find \(P(A), P(B), P(A \cup B),\) and \(P(A \cap B)\) by summing the probabilities of the sample points in these events.
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