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Chapter 2: Problem 60

Components of a certain type are shipped to a supplier in batches of ten. Suppose that \(50 \%\) of all such batches contain no defective components, \(30 \%\) contain one defective component, and \(20 \%\) contain two defective components. Two components from a batch are randomly selected and tested. What are the probabilities associated with 0,1 , and 2 defective components being in the batch under each of the following conditions? a. Neither tested component is defective. b. One of the two tested components is defective. [Hint: Draw a tree diagram with three first-generation branches for the three different types of batches.]

Short Answer

Expert verified
Neither defective: 0 defects ~67%, 1 defect ~25%, 2 defects ~8%. One defective: 0 defects 0%, 1 defect 60%, 2 defects 40%.

Step by step solution

01

Understanding the Problem

We need to calculate the probability of 0, 1, or 2 defective components in a batch. Three batch types exist based on defective components: no defects, one defect, and two defects. We are given the probabilities for these: \( P(A) = 0.5 \) (0 defects), \( P(B) = 0.3 \) (1 defect), and \( P(C) = 0.2 \) (2 defects).
02

Outline of Solution Strategy

For each condition (0 or 1 defective component in the tested pair), compute conditional probabilities for batch types (0, 1, or 2 defectives). Use Bayes' Theorem for calculations.
03

Analyzing Condition A: Neither Tested Component is Defective

For condition a, calculate \( P(A | D_0) \), \( P(B | D_0) \), and \( P(C | D_0) \), where \( D_0 \) signifies neither component is defective. From each batch type, calculate the probability that neither tested component is defective: - \( P(D_0 | A) = 1 \) (0 defects in the batch, so definitely no defects tested)- \( P(D_0 | B) = \frac{8}{10} \times \frac{7}{9} = \frac{56}{90} \) (1 defect, probability both are from among 9 non-defective)- \( P(D_0 | C) = \frac{8}{10} \times \frac{7}{9} = \frac{28}{90} \) (2 defects, keeping probability calculation same for both).
04

Calculate Total Probability of Condition A

Calculate \( P(D_0) = P(D_0 | A)P(A) + P(D_0 | B)P(B) + P(D_0 | C)P(C) \):\[ P(D_0) = 1 \times 0.5 + \frac{56}{90} \times 0.3 + \frac{28}{90} \times 0.2 \] Simplify to get the overall probability for no defective: \[ P(D_0) = 0.5 + 0.1867 + 0.0622 \approx 0.7489 \].
05

Conditional Probabilities for 0 Defective Components in Sample

Use Bayes' theorem to find:- \( P(A|D_0) = \frac{P(D_0|A)P(A)}{P(D_0)} = \frac{0.5}{0.7489} \approx 0.6677 \)- \( P(B|D_0) = \frac{P(D_0|B)P(B)}{P(D_0)} = \frac{0.1867}{0.7489} \approx 0.2493 \)- \( P(C|D_0) = \frac{P(D_0|C)P(C)}{P(D_0)} = \frac{0.0622}{0.7489} \approx 0.083 \).
06

Analyzing Condition B: One Tested Component is Defective

For condition b, calculate \( P(A | D_1) \), \( P(B | D_1) \), and \( P(C | D_1) \), where \( D_1 \) is one defective from selected batch. Probabilities:- \( P(D_1 | A) = 0 \) (0 defects in batch, impossible)- \( P(D_1 | B) = \frac{2}{10} \times \frac{8}{9} \times 2 = \frac{32}{90} \) (probability one defective is chosen)- \( P(D_1 | C) = \left( \frac{2}{10} \times \frac{8}{9} \right) + \left(\frac{8}{10} \times \frac{2}{9}\right) = \frac{32}{90} \) (probability one of two possible defects chosen).
07

Calculate Total Probability of Condition B

Calculate \( P(D_1) = P(D_1 | A)P(A) + P(D_1 | B)P(B) + P(D_1 | C)P(C) \):\[ P(D_1) = 0 + \frac{32}{90} \times 0.3 + \frac{32}{90} \times 0.2 \] Simplify the equation:\[ P(D_1) = 0.1067 + 0.0711 = 0.1778 \].
08

Conditional Probabilities for 1 Defective Component in Sample

Use Bayes' theorem again:- \( P(A|D_1) = \frac{P(D_1|A)P(A)}{P(D_1)} = 0 \) - \( P(B|D_1) = \frac{P(D_1|B)P(B)}{P(D_1)} = \frac{0.1067}{0.1778} \approx 0.6 \)- \( P(C|D_1) = \frac{P(D_1|C)P(C)}{P(D_1)} = \frac{0.0711}{0.1778} \approx 0.4 \).

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Key Concepts

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

Conditional Probability
Conditional probability helps us determine the likelihood of an event occurring given that another event has occurred. It quantifies our intuition that probabilities can change depending on new information. For example, knowing some components of a batch are defective can influence our probability calculations about other components.

Mathematically, conditional probability is defined as the probability of event A occurring given that event B has occurred, indicated by \(P(A|B)\). The formula is:
  • \(P(A|B) = \frac{P(A \cap B)}{P(B)}\)
This equation shows how the intersection of two events affects one particular event's outcome. In the context of defective components, if we know one component is defective, it alters our calculation on the probability of others being defective.
Bayes' Theorem
Bayes' Theorem is a powerful tool in probability theory that lets us update probabilities based on new evidence. It links conditional probabilities and the probabilities of their events.

The theorem is represented as:
  • \(P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}\)
Here, \(P(B|A)\) is the probability of observing evidence B given event A is true, \(P(A)\) is the prior probability of event A, and \(P(B)\) is the probability of evidence B. The theorem allows us to update \(P(A|B)\) as new data (event B) comes in.

In defective components analysis, Bayes' Theorem helps determine the probability of a batch type given test results of components, allowing us to adjust our prior beliefs in light of new evidence.
Defective Components Analysis
In the real world, particularly in manufacturing, analyzing the probability of defective components is crucial for quality control. Defective components analysis involves using probability theory to assess and manage the likelihood of parts being defective within a batch, thereby ensuring product quality and reliability.

This specific analysis often utilizes tree diagrams to visualize and calculate probabilities of different scenarios involving defects. It is essential in identifying the defect ratio in a batch and planning testing strategies. For instance, determining the probability distributions for batches with different numbers of defects helps in drawing test samples, predicting defect occurrences, and informing corrective actions.

Such analyses are instrumental in maintaining high manufacturing standards and minimizing defects, ultimately reducing waste and increasing customer satisfaction through reliable product quality.

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Most popular questions from this chapter

An academic department with five faculty membersAnderson, Box, Cox, Cramer, and Fisher-must select two of its members to serve on a personnel review committee. Because the work will be time-consuming, no one is anxious to serve, so it is decided that the representative will be selected by putting five slips of paper in a box, mixing them, and selecting two. a. What is the probability that both Anderson and Box will be selected? [Hint: List the equally likely outcomes.] b. What is the probability that at least one of the two members whose name begins with \(C\) is selected? c. If the five faculty members have taught for \(3,6,7,10\), and 14 years, respectively, at the university, what is the probability that the two chosen representatives have at least 15 years' teaching experience at the university?

At a certain gas station, \(40 \%\) of the castomers use regular gas \(\left(A_{1}\right), 35 \%\) use plus gas \(\left(A_{2}\right)\), and \(25 \%\) use premium \(\left(A_{3}\right)\). Of those customers using regular gas, only \(30 \%\) fill their tanks (event \(B\) ). Of those customers using plus, \(60 \%\) fill their tanks, whereas of those using premium, \(50 \%\) fill their tanks. a. What is the probability that the next customer will request plus gas and fill the tank \(\left(A_{2} \cap B\right)\) ? b. What is the probability that the next customer fills the tank? c. If the next customer fills the tank, what is the probability that regular gas is requested? Plus? Premium?

a. A lumber company has just taken delivery on a lot of \(10,0002 \times 4\) boards. Suppose that \(20 \%\) of these boards \((2,000)\) are actually too green to be used in first-quality construction. Two boards are selected at random, one after the other. Let \(A=\) [the first board is green] and \(B=\) [the second board is green ). Compute \(P(A), P(B)\), and \(P(A \cap B)\) (a tree diagram might help). Are \(A\) and \(B\) independent? b. With \(A\) and \(B\) independent and \(P(A)=P(B)=.2\), what is \(P(A \cap B)\) ? How much difference is there between this answer and \(P(A \cap B)\) in part (a)? For purposes of calculating \(P(A \cap B)\), can we assume that \(A\) and \(B\) of part (a) are independent to obtain essentially the correct probability? c. Suppose the lot consists of ten boards, of which two are green. Does the assumption of independence now yield approximately the correct answer for \(P(A \cap B)\) ? What is the critical difference between the situation here and that of part (a)? When do you think that an independence assumption would be valid in obtaining an approximately correct answer to \(P(A \cap B)\) ?

A certain system can experience three different types of defects. Let \(A_{i}(i=1,2,3)\) denote the event that the system has a defect of type \(i\). Suppose that $$ \begin{aligned} &P\left(A_{1}\right)=.12 \quad P\left(A_{2}\right)=.07 \quad P\left(A_{3}\right)=.05 \\ &P\left(A_{1} \cup A_{2}\right)=.13 \quad P\left(A_{1} \cup A_{3}\right)=.14 \\\ &P\left(A_{2} \cup A_{3}\right)=.10 \quad P\left(A_{1} \cap A_{2} \cap A_{3}\right)=.01 \end{aligned} $$ a. What is the probability that the system does not have a type 1 defect? b. What is the probability that the system has both type 1 and type 2 defects? c. What is the probability that the system has both type 1 and type 2 defects but not a type 3 defect? d. What is the probability that the system has at most two of these defects?

Let \(A\) denote the event that the next request for assistance from a statistical software consultant relates to the SPSS package, and let \(B\) be the event that the next request is for help with SAS. Suppose that \(P(A)=.30\) and \(P(B)=.50\). a. Why is it not the case that \(P(A)+P(B)=1\) ? b. Calculate \(P\left(A^{\prime}\right)\). c. Calculate \(P(A \cup B)\). d. Calculate \(P\left(A^{\prime} \cap B^{\prime}\right)\).
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