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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?

Short Answer

Expert verified
a. 0.88 b. 0.06 c. 0.05 d. 0.99

Step by step solution

01

Calculate probability of no type 1 defect

Given that \[P(A_1) = 0.12\]the probability that the system does not have a type 1 defect is \[P(A_1^c) = 1 - P(A_1) = 1 - 0.12 = 0.88\]
02

Calculate probability of both type 1 and type 2 defects

The probability of the union of events \(A_1\) and \(A_2\) is given as\[P(A_1 \cup A_2) = 0.13\]Using the principle of inclusion-exclusion, the probability of the intersection of \(A_1\) and \(A_2\) is\[P(A_1 \cap A_2) = P(A_1) + P(A_2) - P(A_1 \cup A_2) = 0.12 + 0.07 - 0.13 = 0.06\]
03

Calculate probability of types 1 and 2 but not type 3 defect

From Step 2, we have\[P(A_1 \cap A_2) = 0.06\]We also know from the problem statement\[P(A_1 \cap A_2 \cap A_3) = 0.01\]The probability of having both type 1 and type 2 defects but not type 3 defect is then\[P(A_1 \cap A_2 \cap A_3^c) = P(A_1 \cap A_2) - P(A_1 \cap A_2 \cap A_3) = 0.06 - 0.01 = 0.05\]
04

Calculate probability of at most two defects

Using the complement principle, the probability of having at most two defects is the complement of having all three defects. We know\[P(A_1 \cap A_2 \cap A_3) = 0.01\]Thus, the probability that the system has at most two defects is\[1 - P(A_1 \cap A_2 \cap A_3) = 1 - 0.01 = 0.99\]

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Key Concepts

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

Inclusion-Exclusion Principle
The Inclusion-Exclusion Principle is a fundamental concept in probability and set theory that helps us find the probability of the union of multiple events. In simple terms, it prevents double-counting when dealing with the overlap of two or more events. Suppose you have two events, A and B. The probability of either A or B occurring, or both, is calculated by adding the probabilities of the individual events and subtracting the probability of their intersection. Mathematically, it is represented as: \[\]\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] \[\]This formula can extend to three or more events, providing a way to include or exclude overlaps sequentially. For instance, when dealing with three events A, B, and C, the principle becomes: \[\]\[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) \] \[\]In the context of our exercise, we used the Inclusion-Exclusion Principle to find the probability of the intersection of events, which involved looking at the combined event of types 1 and 2 defects.
Set Theory
Set Theory is the mathematical study of collections of objects, known as sets. In probability, sets often represent events, and understanding their operations is crucial for solving probability problems. This field employs concepts such as union, intersection, and complement to describe relationships between events.
  • Union (\(A \cup B\)): Represents any element that is in set A, set B, or both.
  • Intersection (\(A \cap B\)): Represents elements common to both sets A and B.
  • Complement (\(A^c\)): Represents elements not in set A.
Set theory principles allow us to visualize and understand different combinations of event outcomes using diagrams like Venn diagrams, which illustrate elements shared among sets or unique to each set. These fundamental definitions are essential for working with probability, such as when assessing intersections and unions to calculate probabilities accurately. In the exercise provided, set operations help detail the interaction between different defect types.
Complementary Events
Complementary events concept is a straightforward yet powerful tool in probability. It revolves around the idea that for any event A, the event that A does not happen is called its complement, denoted as \(A^c\). The probabilities of an event and its complement always add up to 1. This relationship stems from the fact that something either happens or it does not. Mathematically, this is expressed as: \[\]\[ P(A) + P(A^c) = 1 \] \[\]Finding the probability of a complementary event is often simpler and provides insight into related problems. This scenario was tackled in Step 1 of our original exercise, where the probability of not having a type 1 defect was calculated. Understanding and applying the concept of complementary events can significantly streamline solving probability problems and is a foundational tool in probability theory. In many problems, especially those involving multiple possibilities, calculating complementary probabilities aids in evaluating complex scenarios by working with their negations.

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Most popular questions from this chapter

A certain shop repairs both audio and video components. Let \(A\) denote the event that the next component brought in for repair is an audio component, and let \(B\) be the event that the next component is a compact disc player (so the event \(B\) is contained in \(A\) ). Suppose that \(P(A)=.6\) and \(P(B)=.05\). What is \(P(B \mid A)\) ?

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A consulting firm presently has bids out on three projects. Let \(A_{i}=\\{\) awarded project \(i\\}\), for \(i=\) \(1,2,3\), and suppose that \(P\left(A_{1}\right)=.22, P\left(A_{2}\right)=.25\), \(P\left(A_{3}\right)=.28, P\left(A_{1} \cap A_{2}\right)=.11, P\left(A_{1} \cap A_{3}\right)=.05\), \(P\left(A_{2} \cap A_{3}\right)=.07, P\left(A_{1} \cap A_{2} \cap A_{3}\right)=.01\). Express in words each of the following events, and compute the probability of each event: a. \(A_{1} \cup A_{2}\) b. \(A_{1}{ }^{\prime} \cap A_{2}{ }^{\prime} \quad\left[\right.\) Hint : \(\left.\left(A_{1} \cup A_{2}\right)^{\prime}=A_{1}{ }^{\prime} \cap A_{2}{ }^{\prime}\right]\) c. \(A_{1} \cup A_{2} \cup A_{3}\) d. \(A_{1}{ }^{\prime} \cap A_{2}{ }^{\prime} \cap A_{3}{ }^{\prime}\) e. \(A_{1}{ }^{\prime} \cap A_{2}{ }^{\prime} \cap A_{3}\) f. \(\left(A_{1}^{\prime} \cap A_{2}^{\prime}\right) \cup A_{3}\)

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