/*! 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} Problem 15 A consulting firm presently has ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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. \(\left(A_{1}^{\prime} \cap A_{2}^{\prime}\right) \cup A_{3}\)

Short Answer

Expert verified
a. 0.36, b. 0.64, c. 0.53, d. 0.47, e. 0.1792, f. 0.7408.

Step by step solution

01

Understand the Meaning of Each Event

We are given events related to three projects where project \( A_i \) corresponds to being awarded the project \( i \). The complementary event \( A_i^{\prime} \) means not being awarded the project \( i \).- \( A_1 \cup A_2 \) means being awarded either project 1 or project 2 or both.- \( A_1^{\prime} \cap A_2^{\prime} \) means neither project 1 nor project 2 is awarded.- \( A_1 \cup A_2 \cup A_3 \) means being awarded at least one of the three projects.- \( A_1^{\prime} \cap A_2^{\prime} \cap A_3^{\prime} \) means none of the projects is awarded.- \( A_1^{\prime} \cap A_2^{\prime} \cap A_3 \) means only project 3 is awarded.- \( (A_1^{\prime} \cap A_2^{\prime}) \cup A_3 \) means either neither project 1 nor project 2 is awarded or project 3 is awarded.
02

Calculate Probability of \( A_1 \cup A_2 \)

Use the formula for the union of two events:\[ P(A_1 \cup A_2) = P(A_1) + P(A_2) - P(A_1 \cap A_2) \]Substitute the given values:\[ P(A_1 \cup A_2) = 0.22 + 0.25 - 0.11 = 0.36 \]
03

Calculate Probability of \( A_1^{\prime} \cap A_2^{\prime} \)

Since this is the complement of \( A_1 \cup A_2 \), use the formula for the complement:\[ P(A_1^{\prime} \cap A_2^{\prime}) = 1 - P(A_1 \cup A_2) \]Substitute the value from Step 2:\[ P(A_1^{\prime} \cap A_2^{\prime}) = 1 - 0.36 = 0.64 \]
04

Calculate Probability of \( A_1 \cup A_2 \cup A_3 \)

Use the formula for the union of three events:\[P(A_1 \cup A_2 \cup A_3) = P(A_1) + P(A_2) + P(A_3)\ - P(A_1 \cap A_2) - P(A_2 \cap A_3) - P(A_1 \cap A_3)\ + P(A_1 \cap A_2 \cap A_3) \]Substitute the given values:\[ P(A_1 \cup A_2 \cup A_3) = 0.22 + 0.25 + 0.28 - 0.11 - 0.07 - 0.05 + 0.01 = 0.53 \]
05

Calculate Probability of \( A_1^{\prime} \cap A_2^{\prime} \cap A_3^{\prime} \)

Since this is the complement of \( A_1 \cup A_2 \cup A_3 \), use the complement formula:\[ P(A_1^{\prime} \cap A_2^{\prime} \cap A_3^{\prime}) = 1 - P(A_1 \cup A_2 \cup A_3) \]Substitute the value from Step 4:\[ P(A_1^{\prime} \cap A_2^{\prime} \cap A_3^{\prime}) = 1 - 0.53 = 0.47 \]
06

Calculate Probability of \( A_1^{\prime} \cap A_2^{\prime} \cap A_3 \)

We find this by considering no awards for projects 1 and 2, but project 3 is awarded:\[ P(A_1^{\prime} \cap A_2^{\prime} \cap A_3) = P(A_1^{\prime} \cap A_2^{\prime}) \cdot P(A_3|A_1^{\prime} \cap A_2^{\prime}) \]Since no conditions are specified, assume independence:\[ P(A_1^{\prime} \cap A_2^{\prime} \cap A_3) = 0.64 \cdot 0.28 = 0.1792 \]
07

Calculate Probability of \((A_1^{\prime} \cap A_2^{\prime}) \cup A_3\)

Use the formula for the union of two events:\[ P((A_1^{\prime} \cap A_2^{\prime}) \cup A_3) = P(A_1^{\prime} \cap A_2^{\prime}) + P(A_3) - P((A_1^{\prime} \cap A_2^{\prime}) \cap A_3) \]Substitute the values:\[ P((A_1^{\prime} \cap A_2^{\prime}) \cup A_3) = 0.64 + 0.28 - 0.1792 = 0.7408 \]

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 91Ó°ÊÓ!

Key Concepts

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

Event Probability Calculation
Calculating the probability of an event is a fundamental concept within probability theory. Understanding it allows you to quantify the chance of an event happening. An event in probability usually relates to some specific outcome, like flipping a coin and getting heads. In problems involving multiple events, we often have to determine the probability of compounded or combined events occurring. For example, if we have events related to projects, such as being awarded specific projects, we may want to know the probability of being awarded at least one project. We do this by using other concepts, like the union and intersection of events in probability. To calculate the probability of an event properly, you need to follow these steps:
  • Identify the probabilities of individual events, such as the probability of each project being awarded.
  • Use probability rules, such as the additive rule for mutually exclusive events and formulas incorporating intersections for non-mutually exclusive events.
  • Pay attention to any given conditional probabilities and understand whether the events are considered independent or dependent.
Through these steps, you can accurately calculate the probability of any event or combination of events, which is a pivotal skill in probability theory.
Complementary Events
Complementary events are a vital concept in probability, helping you to understand the complete range of possible outcomes.A complementary event is one that reveals the probability of an event not happening. If an event is defined as "A," its complement, denoted as "A'", is the set of all single outcomes in the sample space not in "A."For instance, if the probability of an event, such as being awarded a project, is 0.22, the probability of not being awarded the project, its complement, would be 0.78. The formula to find the probability of complementary events is:\[ P(A') = 1 - P(A) \]This formula is powerful because it allows you to deduce one probability from the other, making it especially handy when calculating probabilities of more complex events. It simplifies when you're looking for the chance of none of a set of events occurring, often writing this as the complement of their union.Complementary events are a building block for understanding how probabilities are assigned across all possible outcomes in a probability space.
Union and Intersection of Events
Union and intersection are operations that help in handling probabilities of multiple events occurring together or separately.**Union of Events**The union operation, depicted as \(A \cup B\), refers to the scenario where either event A occurs, event B occurs, or both occur. This is akin to asking if at least one event happens.For example, calculating the probability \(P(A_1 \cup A_2)\) would involve adding probabilities of \(A_1\) and \(A_2\) and subtracting the intersection since it would be double-counted.
**Intersection of Events**The intersection, denoted \(A \cap B\), is about both events happening simultaneously. Its probability, \(P(A \cap B)\), gives the likelihood that both events occur at the same time.Calculating this requires more information about the relationship between the events: whether they're independent or not. For independent events, you multiply their individual probabilities.
Using these operations and understanding how to blend both different and overlapping event occurrences guides you to correctly assess complex situations. It's the strategic use of them that makes probability theory both a powerful analytic tool and a fascinating study subject.

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

A particular iPod playlist contains 100 songs, of which 10 are by the Beatles. Suppose the shuffle feature is used to play the songs in random order (the randomness of the shuffling process is investigated in "Does Your iPod Really Play Favorites?" (The Amer. Statistician, 2009: 263 - 268)). What is the probability that the first Beatles song heard is the fifth song played?

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 indepth 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?

The computers of six faculty members in a certain department are to be replaced. Two of the faculty members have selected laptop machines and the other four have chosen desktop machines. Suppose that only two of the setups can be done on a particular day, and the two computers to be set up are randomly selected from the six (implying 15 equally likely outcomes; if the computers are numbered \(1,2, \ldots, 6\), then one outcome consists of computers 1 and 2, another consists of computers 1 and 3 , and so on). a. What is the probability that both selected setups are for laptop computers? b. What is the probability that both selected setups are desktop machines? c. What is the probability that at least one selected setup is for a desktop computer? d. What is the probability that at least one computer of each type is chosen for setup?

Three molecules of type \(A\), three of type \(B\), three of type \(C\), and three of type \(D\) are to be linked together to form a chain molecule. One such chain molecule is \(A B C D A B C D A B C D\), and another is \(B C D D A A A B D B C C\). a. How many such chain molecules are there? [Hint: If the three A's were distinguishable from one another- \(A_{1}, A_{2}, A_{3}\)-and the \(B\) 's, \(C\) 's, and \(D\) 's were also, how many molecules would there be? How is this number reduced when the subscripts are removed from the \(A\) 's?] b. Suppose a chain molecule of the type described is randomly selected. What is the probability that all three molecules of each type end up next to each other (such as in \(B B B A A A D D D C C C\) )?

Use the axioms to show that if one event \(A\) is contained in another event \(B\) (i.e., \(A\) is a subset of \(B\) ), then \(P(A) \leq P(B)\). [Hint: For such \(A\) and \(B, A\) and \(B \cap A^{\prime}\) are disjoint and \(B=A \cup\left(B \cap A^{\prime}\right)\), as can be seen from a Venn diagram.] For general \(A\) and \(B\), what does this imply about the relationship among \(P(A \cap B), P(A)\), and \(P(A \cup B)\) ?

See all solutions

Recommended explanations on Math Textbooks

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.