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Chapter 2: Problem 17

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 \(P\left(A^{\prime} \cap B^{\prime}\right)\).

Short Answer

Expert verified
a. A and B are not the only possible events. b. 0.70 c. 0.80 d. 0.20.

Step by step solution

01

Analyze Event Probabilities

The probability of the occurrence of two events \(A\) and \(B\) are given as \(P(A) = 0.30\) and \(P(B) = 0.50\). The sum \(P(A) + P(B) = 0.80\) does not equal to 1. This indicates that \(A\) and \(B\) are not the only possible events and there is the possibility that another event \(C\) (such as the request being for another software or no request at all) can occur.
02

Calculate the Complement Probability of A

The complement of event \(A\), denoted \(A'\), is the event that the next request is not for the SPSS package. By definition, \(P(A') = 1 - P(A)\). Thus, \(P(A') = 1 - 0.30 = 0.70\).
03

Calculate the Union Probability

The probability of either event \(A\) or event \(B\) occurring, denoted \(P(A \cup B)\), can be calculated using the formula \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). Since \(A\) and \(B\) are not mentioned as mutually exclusive, to find \(P(A \cup B)\), we assume they may overlap. Without specific information, we cannot include \(P(A \cap B)\). Hence, we use the max assumption with no overlap: \(P(A \cup B) = 0.30 + 0.50 = 0.80\).
04

Calculate the Probability of Neither A nor B

The event \(A' \cap B'\) is the occurrence of neither \(A\) nor \(B\). This is the complement of \(A \cup B\). Therefore, \(P(A' \cap B') = 1 - P(A \cup B)\). From Step 3, \(P(A \cup B) = 0.80\). Thus, \(P(A' \cap B') = 1 - 0.80 = 0.20\).

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Key Concepts

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

Event Probability
Event probability refers to the chance that a specific event will happen. In probability theory, an event is a set of outcomes of an experiment. For instance, in the given exercise, event \( A \) is about assistance requests for the SPSS package, while event \( B \) is for SAS help. Probabilities are numbers between 0 and 1, indicating the likelihood from impossible (0) to definite (1).

When the exercise mentions that \( P(A) = 0.30 \) and \( P(B) = 0.50 \), it means that there is a 30% chance a request is for SPSS and a 50% chance it is for SAS. Importantly, the sum \( P(A) + P(B) = 0.80 \) is less than 1, highlighting that these are not the only possibilities! Perhaps there are requests for different software, or perhaps no requests at all. These options combined with events \( A \) and \( B \) make up all potential outcomes.
Complementary Events
Complementary events are those that are mutually exclusive and collectively exhaustive, meaning they cover all possible outcomes of a probability scenario without overlapping. The complement of an event \( A \), usually denoted as \( A' \), is the event that \( A \) does not happen. Simply put, it's whatever remains when \( A \) doesn't occur.

To find the probability of the complement of an event, you use the formula \( P(A') = 1 - P(A) \). Here, in the case of SPSS requests, \( A' \) involves situations where the request isn't for SPSS: \( P(A') = 1 - 0.30 = 0.70 \). This tells us that there's a 70% probability that the next request will be about something other than SPSS. Remember that complements help fill in the picture by outlining what doesn't happen alongside what does.
Union of Events
In probability, the concept of "union" ( denoted as \( A \cup B \) ) is all about the likelihood of either event \( A \), \( B \), or both occurring. It represents the combined probability that at least one of the involved events will happen.

The formula used to calculate the union of two events is \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \), where \( P(A \cap B) \) is the probability of both events occurring simultaneously. In the absence of this overlap information, because events \( A \) and \( B \) were not specified to be mutually exclusive, we approximate that they do not interfere. Therefore, \( P(A \cup B) = 0.30 + 0.50 = 0.80 \). This means there's an 80% chance that we will receive a request for either SPSS or SAS assistance. Understanding union helps in grasping how different probabilities interact and often grow together.

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Most popular questions from this chapter

A system consists of two components. The probability that the second component functions in a satisfactory manner during its design life is \(.9\), the probability that at least one of the two components does so is \(.96\), and the probability that both components do so is .75. Given that the first component functions in a satisfactory manner throughout its design life, what is the probability that the second one does also?

There has been a great deal of controversy over the last several years regarding what types of surveillance are appropriate to prevent terrorism. Suppose a particular surveillance system has a \(99 \%\) chance of correctly identifying a future terrorist and a \(99.9 \%\) chance of correctly identifying someone who is not a future terrorist. If there are 1000 future terrorists in a population of 300 million, and one of these 300 million is randomly selected, scrutinized by the system, and identified as a future terrorist, what is the probability that he/she actually is a future terrorist? Does the value of this probability make you uneasy about using the surveillance system? Explain.

Consider randomly selecting a student at a certain university, and let \(A\) denote the event that the selected individual has a Visa credit card and \(B\) be the analogous event for a MasterCard. Suppose that \(P(A)=.5, P(B)=.4\), and \(P(A \cap B)=.25\). a. Compute the probability that the selected individual has at least one of the two types of cards (i.e., the probability of the event \(A \cup B\) ). b. What is the probability that the selected individual has neither type of card? c. Describe, in terms of \(A\) and \(B\), the event that the selected student has a Visa card but not a MasterCard, and then calculate the probability of this event.

A company uses three different assembly lines- \(A_{1}, A_{2}\), and \(A_{3}\)-to manufacture a particular component. Of those manufactured by line \(A_{1}, 5 \%\) need rework to remedy a defect, whereas \(8 \%\) of \(A_{2}\) 's components need rework and \(10 \%\) of \(A_{3}\) 's need rework. Suppose that \(50 \%\) of all components are produced by line \(A_{1}, 30 \%\) are produced by line \(A_{2}\), and \(20 \%\) come from line \(A_{3}\). If a randomly selected component needs rework, what is the probability that it came from line \(A_{1}\) ? From line \(A_{2}\) ? From line \(A_{3}\) ?

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 in-depth 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?
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