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Chapter 2: Problem 13

A computer 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. \(A_{1}^{\prime} \cap A_{2}^{\prime}\left[\right.\) Hinf: \(\left.\left(A_{1} \cup A_{2}\right)^{\prime}=A_{1}^{\prime} \cap A_{2}^{\prime}\right]\) 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. 0.17 f. 0.75

Step by step solution

01

Calculate Probability of Event a

Event a is expressed as \(A_{1} \cup A_{2}\), which means the probability that the firm is awarded either project 1 or project 2, or both. To find the probability of the union of two events, use the formula:\[P(A_{1} \cup A_{2}) = P(A_{1}) + P(A_{2}) - P(A_{1} \cap A_{2})\]Substitute the given probabilities:\[P(A_{1} \cup A_{2}) = 0.22 + 0.25 - 0.11 = 0.36\]
02

Calculate Probability of Event b

Event b is expressed as \(A_{1}^{\prime} \cap A_{2}^{\prime}\). This is the probability that the firm is not awarded both projects 1 and 2. From the hint, \( (A_{1} \cup A_{2})^{\prime} = A_{1}^{\prime} \cap A_{2}^{\prime}\), so we use the complement rule:\[P(A_{1}^{\prime} \cap A_{2}^{\prime}) = 1 - P(A_{1} \cup A_{2})\]Using our result from step 1:\[P(A_{1}^{\prime} \cap A_{2}^{\prime}) = 1 - 0.36 = 0.64\]
03

Calculate Probability of Event c

Event c is expressed as \(A_{1} \cup A_{2} \cup A_{3}\), which refers to the probability that the firm is awarded at least one of the three projects. Use the formula for the union of three events:\[P(A_{1} \cup A_{2} \cup A_{3}) = P(A_{1}) + P(A_{2}) + P(A_{3}) - P(A_{1} \cap A_{2}) - P(A_{1} \cap A_{3}) - P(A_{2} \cap A_{3}) + P(A_{1} \cap A_{2} \cap A_{3})\]Substitute the given values:\[P(A_{1} \cup A_{2} \cup A_{3}) = 0.22 + 0.25 + 0.28 - 0.11 - 0.05 - 0.07 + 0.01 = 0.53\]
04

Calculate Probability of Event d

Event d is expressed as \(A_{1}^{\prime} \cap A_{2}^{\prime} \cap A_{3}^{\prime}\), which represents the probability that none of the projects are awarded to the firm. Recognize this as the complement of \(A_{1} \cup A_{2} \cup A_{3}\):\[P(A_{1}^{\prime} \cap A_{2}^{\prime} \cap A_{3}^{\prime}) = 1 - P(A_{1} \cup A_{2} \cup A_{3})\]Using the result from step 3:\[P(A_{1}^{\prime} \cap A_{2}^{\prime} \cap A_{3}^{\prime}) = 1 - 0.53 = 0.47\]
05

Calculate Probability of Event e

Event e is expressed as \(A_{1}^{\prime} \cap A_{2}^{\prime} \cap A_{3}\), which refers to the probability that neither project 1 nor project 2 is awarded but project 3 is. By using complement rules and known results:\[P(A_{1}^{\prime} \cap A_{2}^{\prime} \cap A_{3}) = P(A_{3}) - P(A_{1} \cap A_{3}) - P(A_{2} \cap A_{3}) + P(A_{1} \cap A_{2} \cap A_{3})\]Substitute the given probabilities:\[P(A_{1}^{\prime} \cap A_{2}^{\prime} \cap A_{3}) = 0.28 - 0.05 - 0.07 + 0.01 = 0.17\]
06

Calculate Probability of Event f

Event f is expressed as \((A_{1}^{\prime} \cap A_{2}^{\prime}) \cup A_{3}\), which refers to the probability where projects 1 and 2 are not awarded but project 3 might be. Use the known values:\[P((A_{1}^{\prime} \cap A_{2}^{\prime}) \cup A_{3}) = P(A_{1}^{\prime} \cap A_{2}^{\prime}) + P(A_{3}) - P(A_{1}^{\prime} \cap A_{2}^{\prime} \cap A_{3})\]From previous steps:\[P((A_{1}^{\prime} \cap A_{2}^{\prime}) \cup A_{3}) = 0.64 + 0.28 - 0.17 = 0.75\]

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Key Concepts

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

Union of Events
In probability theory, the union of events is a concept that helps us determine the probability of at least one of several events occurring. Let's delve deeper into this concept using the example from the problem provided. We define the events as follows:
  • \(A_1\): The firm is awarded project 1.
  • \(A_2\): The firm is awarded project 2.
  • \(A_3\): The firm is awarded project 3.
The union of events \(A_1 \cup A_2\) is essentially the probability that the firm is awarded either project 1, project 2, or both, so at least one of the two projects. The formula to find this probability is: \[P(A_1 \cup A_2) = P(A_1) + P(A_2) - P(A_1 \cap A_2)\]The term \(P(A_1 \cap A_2)\) is subtracted because it is counted twice when adding \(P(A_1)\) and \(P(A_2)\), so we must remove the overlap. This formula can be generalized for any number of events, allowing us to calculate the probability of at least one of multiple projects being awarded. It's a key principle in probability that underscores the inclusivity of an 'or' situation.
Complement Rule
The complement rule is a powerful tool to find the probability of the occurrence of an event not happening. It uses the fact that probabilities of all possible outcomes sum up to 1. This rule is helpful to calculate probabilities of complex events by examining what doesn't occur.Using the example from the exercise, consider the probability of not being awarded either project 1 or project 2, denoted by \(A_1' \cap A_2'\). According to the complement rule:
  • \(P(A_1' \cap A_2') = 1 - P(A_1 \cup A_2)\)
Here, \(A_1'\) denotes the event that the firm is not awarded project 1, and similarly, \(A_2'\) denotes not being awarded project 2. The formula tells us that the probability of these 'not' events can be derived directly from the probability of their respective 'do' events, enhancing our ability to conduct efficient probability calculations without examining each event separately. This method also scales well for calculating the complement across multiple events, providing a simple yet effective strategy.
Intersection of Events
The intersection of events in probability but also in this exercise represents the scenario where multiple events happen simultaneously. The focus is on calculating the probability that more than one event occurs at the same time.For example, the probability of the firm being awarded both projects 1 and 2 is represented as \(P(A_1 \cap A_2)\). This is the probability that both events occur, not separately but together. To find such probabilities, it's often helpful directly from data provided or sometimes using specific probability rules. In our exercise:
  • \(P(A_1 \cap A_2) = 0.11\), showing that both projects 1 and 2 are awarded simultaneously.
When tackling multiple intersections, especially with more projects, we use the intersection rule of three events: \[P(A_1 \cap A_2 \cap A_3)\]Understanding intersections is crucial because they help determine shared outcomes across different events. They are the connective tissues in probability that help gauge concurrent outcomes in real-world scenarios.

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Most popular questions from this chapter

In five-card poker, a straight consists of five cards with adjacent denominations (e.g., 9 of clubs, 10 of hearts, jack of hearts, queen of spades, and king of clubs). Assuming that aces can be high or low, if you are dealt a five-card hand, what is the probability that it will be a straight with high card 10 ? What is the probability that it will be a straight? What is the probability that it will be a straight flush (all cards in the same suit)?

The probability that a grader will make a marking error on any particular question of a multiple-choice exam is .1. If there are ten questions and questions are marked independently, what is the probability that no errors are made? That at least one error is made? If there are \(n\) questions and the probability of a marking error is \(p\) rather than \(.1\), give expressions for these two probabilities.

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