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Chapter 6: Problem 5

A new model of laptop computer can be ordered with one of three screen sizes ( 10 inches, 12 inches, 15 inches) and one of four hard drive sizes \((50 \mathrm{~GB}, 100 \mathrm{~GB}\), \(150 \mathrm{~GB}\), and \(200 \mathrm{~GB}\) ). Consider the chance experiment in which a laptop order is selected and the screen size and hard drive size are recorded. a. Display possible outcomes using a tree diagram. b. Let \(A\) be the event that the order is for a laptop with a screen size of 12 inches or smaller. Let \(B\) be the event that the order is for a laptop with a hard drive size of at most \(100 \mathrm{~GB}\). What outcomes are in \(A^{C}\) ? In \(A \cup B ?\) In \(A \cap B\) ? c. Let \(C\) denote the event that the order is for a laptop with a \(200 \mathrm{~GB}\) hard drive. Are \(A\) and \(C\) disjoint events? Are \(B\) and \(C\) disjoint?

Short Answer

Expert verified
a. 12 outcomes represented in a tree diagram. b. \(A^{C}\): {(15, 50), (15, 100), (15, 150), (15, 200)}. \(A \cup B\): All 12 outcomes. \(A \cap B\): {(10, 50), (10, 100), (12, 50), (12, 100)}. c. A and C are not disjoint; B and C are disjoint.

Step by step solution

01

Construct a Tree Diagram

Start by drawing a tree diagram to represent all possibilities. The first branch indicates the screen size choices (10, 12, 15), and each of these branches further splits into four branches representing the hard drive choices (50, 100, 150, 200). This results in 12 outcomes overall.
02

Consider Event A and Its Complement

Event A consists of orders for laptops with a screen size of 12 inches or smaller, which is (10, 50), (10, 100), (10, 150), (10, 200), (12, 50), (12, 100), (12, 150), (12, 200). The complement of A, \(A^{C}\), consists of the remaining outcomes, which is (15, 50), (15, 100), (15, 150), (15, 200).
03

Consider Events A and B

Event B consists of orders for laptops with a hard drive size of at most 100 GB: (10, 50), (10, 100), (12, 50), (12, 100), (15, 50), (15, 100). The union of A and B, \(A \cup B\), is any outcome in either A or B or both, which includes all 12 outcomes. The intersection of A and B, \(A \cap B\), consists of outcomes present in both A and B, which are (10, 50), (10, 100), (12, 50), (12, 100).
04

Consider Events A, B, and C

Event C consists of orders for laptops with a 200 GB hard drive: (10, 200), (12, 200), (15, 200). A and C share two outcomes (10, 200) and (12, 200), so they are not disjoint. B and C do not share any outcomes, so they are disjoint.

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Key Concepts

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

Probability and Statistics
Probability and statistics are branches of mathematics that deal with data analysis and the likelihood of events occurring. A fundamental concept in this field is the representation of potential outcomes for experiments, which is where tools like tree diagrams come into play.

A tree diagram breaks down complex probability problems into simple, manageable parts. In our exercise, the tree diagram is used to represent all possible choices for screen sizes and hard drive capacities of a new laptop model. Each 'branch' of the diagram corresponds to one decision point, creating a visual representation of all combinations. This helps in making probability calculations more systematic and transparent.

Tree diagrams depict the sequential nature of decisions and their associated probabilities, if any, allowing us to explore and communicate the entire sample space of an experiment efficiently. By visualizing the outcomes, students can often more easily understand the relationship between events, such as the likelihood of ordering a laptop with a specific screen size or hard drive capacity.
Complement of an Event
In probability, the complement of an event is the set of all outcomes in the sample space that are not included in the event. For event A, denoted as 'the order is for a laptop with a screen size of 12 inches or smaller,' the complement is represented as \(A^{C}\) and includes all laptop configurations that do not meet the criteria of event A.

The concept of the complement is crucial because it allows us to calculate the probability of an event not occurring directly. In essence, if we know the probability of an event, we also know the probability of its complement since the sum of these probabilities equals one. In our exercise, we identified the outcomes for event A, thus easily deriving its complement, \(A^{C}\), as the remaining outcomes. Understanding complements is also foundational for grasping more complex concepts like the union and intersection of events.
Union and Intersection of Events
The union of two events, denoted by \( A \cup B \), consists of outcomes that are in either event A or event B or in both. It's the 'either/or' scenario in terms of probability, where we consider all possibilities that satisfy at least one of the event conditions. In our exercise, when we explore \( A \cup B \), we consider all laptops with a screen size of 12 inches or smaller, or with hard drive sizes of at most 100 GB, covering the entire set of outcomes.

Contrastingly, the intersection of two events, denoted by \( A \cap B \), is the set of outcomes that are in both event A and event B simultaneously. It鈥檚 the 'both' scenario where we focus on outcomes that meet both criteria. For the given exercise, \( A \cap B \) includes only laptops that are 12 inches or smaller with hard drives of at most 100 GB. Clarifying these concepts helps students think critically about how different conditions relate to each other and how to calculate probabilities that require multiple conditions to be satisfied at once.

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Most popular questions from this chapter

Many fire stations handle emergency calls for medical assistance as well as calls requesting firefighting equipment. A particular station says that the probability that an incoming call is for medical assistance is .85. This can be expressed as \(P(\) call is for medical assistance \()=.85\). a. Give a relative frequency interpretation of the given probability. b. What is the probability that a call is not for medical assistance? c. Assuming that successive calls are independent of one another, calculate the probability that two successive calls will both be for medical assistance. d. Still assuming independence, calculate the probability that for two successive calls, the first is for medical assistance and the second is not for medical assistance. e. Still assuming independence, calculate the probability that exactly one of the next two calls will be for medical assistance. (Hint: There are two different possibilities. The one call for medical assistance might be the first call, or it might be the second call.) f. Do you think that it is reasonable to assume that. the requests made in successive calls are independent? Explain.

The article "Chances Are You Know Someone with a Tattoo, and He's Not a Sailor" (Associated Press, June 11, 2006) included results from a survey of adults aged 18 to 50 . The accompanying data are consistent with summary values given in the article. $$ \begin{array}{l|cc} & \text { At Least One Tattoo } & \text { No Tattoo } \\ \hline \text { Age 18-29 } & 18 & 32 \\ \text { Age 30-50 } & 6 & 44 \\ \hline \end{array} $$ Assuming these data are representative of adult Americans and that an adult American is selected at random, use the given information to estimate the following probabilities. a. \(P(\) tattoo \()\) b. \(P(\) tattoo \(\mid\) age \(18-29)\) c. \(P(\) tattoo \(\mid\) age \(30-50\) ) d. \(P(\) age \(18-29 \mid\) tattoo \()\)

A medical research team wishes to evaluate two different treatments for a disease. Subjects are selected two at a time, and then one of the pair is assigned to each of the two treatments. The treatments are applied, and each is either a success (S) or a failure (F). The researchers keep track of the total number of successes for each treatment. They plan to continue the chance experiment until the number of successes for one treatment exceeds the number of successes for the other treatment by 2 . For example, they might observe the results in the table for Exercise \(6.82\) given below. The chance experiment would stop after the sixth pair, because Treatment 1 has 2 more successes than Treatment \(2 .\) The researchers would conclude that Treatment 1 is preferable to Treatment \(2 .\) Suppose that Treatment 1 has a success rate of \(.7\) (that is, \(P(\) success \()=.7\) for Treatment 1\()\) and that Treatment 2 has a success rate of \(.4 .\) Use simulation to estimate the probabilities in Parts (a) and (b). (Hint: Use a pair of random digits to simulate one pair of subjects. Let the first digit represent Treatment 1 and use \(1-7\) as an indication of a success and 8,9 , and 0 to indicate a failure. Let the second digit represent Treatment 2, with 1-4 representing a success. For example, if the two digits selected to represent a pair were 8 and 3 , you would record failure for Treatment 1 and success for Treatment 2\. Continue to select pairs, keeping track of the total number of successes for each treatment. Stop the trial as soon as the number of successes for one treatment exceeds that for the other by \(2 .\) This would complete one trial. Now repeat this whole process until you have results for at least 20 trials [more is better]. Finally, use the simulation results to estimate the desired probabilities.) a. Estimate the probability that more than five pairs must be treated before a conclusion can be reached. (Hint: \(P(\) more than 5\()=1-P(5\) or fewer \() .\) ) b. Estimate the probability that the researchers will incorrectly conclude that Treatment 2 is the better treatment.

Only \(0.1 \%\) of the individuals in a certain population have a particular disease (an incidence rate of .001). Of th贸se wh貌 have the disease, \(95 \%\) test possitive wh猫n a certain diagnostic test is applied. Of those who do not have the disease, \(90 \%\) test negative when the test is applied. Suppose that an individual from this population is randomly selected and given the test. a. Construct a tree diagram having two first-generation branches, for has disease and doesn't have disease, and two second-generation branches leading out from each of these, for positive test and negative test. Then enter appropriate probabilities on the four branches. b. Use the general multiplication rule to calculate \(P\) (has disease and positive test). c. Calculate \(P\) (positive test). d. Calculate \(P\) (has disease| positive test). Does the result surprise you? Give an intuitive explanation for the size of this probability.

Five hundred first-year students at a state university were classified according to both high school GPA and whether they were on academic probation at the end of their first semester. The data are $$ \begin{array}{lcccc} && {\text { High School GPA }} \\ & 2.5 \text { to } & 3.0 \text { to } & 3.5 \text { and } & \\ \text { Probation } & <3.0 & <3.5 & \text { Above } & \text { Total } \\ \hline \text { Yes } & 50 & 55 & 30 & 135 \\ \text { No } & 45 & 135 & 185 & 365 \\ \text { Total } & 95 & 190 & 215 & 500 \\ \hline \end{array} $$ a. Construct a table of the estimated probabilities for each GPA-probation combination. b. Use the table constructed in Part (a) to approximate the probability that a randomly selected first-year student at this university will be on academic probation at the end of the first semester. c. What is the estimated probability that a randomly selected first-year student at this university had a high school GPA of \(3.5\) or above? d. Are the two outcomes selected student has a bigh school GPA of \(3.5\) or above and selected student is on academic probation at the end of the first semester independent outcomes? How can you tell? e. Estimate the proportion of first-year students with high school GPAs between \(2.5\) and \(3.0\) who are on academic probation at the end of the first semester. f. Estimate the proportion of those first-year students with high school GPAs \(3.5\) and above who are on academic probation at the end of the first semester.
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