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A large department store offers online ordering. When a purchase is made online, the customer can select one of four different delivery options: expedited overnight delivery, expedited second-business-day delivery, standard delivery, or delivery to the nearest store for customer pick-up. Consider the chance experiment that consists of observing the selected delivery option for a randomly selected online purchase. a. What are the simple events that make up the sample space for this experiment? b. Suppose that the probability of an overnight delivery selection is \(.1\), the probability of a second-day delivery selection is \(.3\), and the probability of a standard-delivery selection is . \(4 .\) Find the following probabilities: i. the probability that a randomly selected online purchase selects delivery to the nearest store for customer pick-up. ii. the probability that the customer selects a form of expedited delivery. iii. the probability that either standard delivery or delivery to the nearest store is selected.

Short Answer

Expert verified
a) The simple events that make up the sample space for this experiment are: expedited overnight delivery, expedited second-business-day delivery, standard delivery, and delivery to the nearest store for customer pick-up. b) i)The probability that a randomly selected online purchase selects delivery to the nearest store for customer pick-up is 0.2. ii) The probability that the customer selects a form of expedited delivery is 0.4. iii) The probability that either standard delivery or delivery to the nearest store is selected is 0.6.

Step by step solution

01

Identify the Sample Space

In this case, the sample space is composed of all possible delivery options a customer can choose when making an online purchase at this department store. These options, which represent our simple events, are: expedited overnight delivery, expedited second-business-day delivery, standard delivery and delivery to the nearest store for customer pick-up.
02

Calculate the Missing Probability

To find the probability of the pick-up delivery, you need to understand that the sum of all probabilities in a sample space always equals 1. Hence, the probability of a pick-up delivery method, \(P(\text{Pick-up})\), is calculated as follows: \(P(\text{Pick-up}) = 1 - P(\text{Overnight}) - P(\text{Second-day}) - P(\text{Standard}) = 1 - .1 - .3 - .4 = .2\). So, 20% of the customers prefer the pick-up method.
03

Calculate the Probability of Expedited Delivery

Here, you add the probabilities of the two events that constitute expedited delivery (overnight and second-business-day delivery). So, \(P(\text{Expedited}) = P(\text{Overnight}) + P(\text{Second-day}) = .1 + .3 = .4\). This means there's a 40% chance that a customer will choose an expedited delivery option.
04

Calculate the Probability of either Standard or Pick-up Delivery

Similarly, you compute the combined probability of two possible outcomes. Thus, \(P(\text{Standard or Pick-up}) = P(\text{Standard}) + P(\text{Pick-up}) = .4 + .2 = .6\). This means there's a 60% probability that a customer will choose either the standard delivery or the pick-up option.

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Key Concepts

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

Sample Space in Probability
Understanding the sample space is foundational in the study of probability. It encapsulates all the potential outcomes of a probability experiment. In the context of our delivery options scenario, where a department store offers four distinct methods for receiving online orders, the sample space consists of these four options: expedited overnight delivery, expedited second-business-day delivery, standard delivery, and delivery to the nearest store for customer pick-up.

Each of these options is known as a 'simple event'—an outcome that cannot be broken down any further. The sample space is important because it sets the stage for determining the likelihood of any combination of events, such as the chance of a customer selecting any specific delivery option. Recognizing the full range of possibilities is crucial when calculating probabilities, which we will explore further in the subsequent sections.
Calculating Probabilities
Calculating probabilities involves determining the chances of various events occurring within our defined sample space. The probabilities of all simple events in a sample space should sum up to 1, representing the certainty that one of the events will occur. Let's take a closer look at the delivery example. Here, we calculated the probability of the pick-up delivery option by subtracting the sum of the known probabilities from 1:

  • \(P(\text{Pick-up}) = 1 - P(\text{Overnight}) - P(\text{Second-day}) - P(\text{Standard}) = 1 - .1 - .3 - .4 = .2\)
This calculation highlights the role of the complementary events in probability. To find the odds of expedited delivery, a composite event, we simply summed the probabilities of the overnight and second-business-day delivery options:

  • \(P(\text{Expedited}) = P(\text{Overnight}) + P(\text{Second-day}) = .1 + .3 = .4\)
Similarly, for a compound event comprising the standard and pick-up delivery options, their individual probabilities were added together:

  • \(P(\text{Standard or Pick-up}) = P(\text{Standard}) + P(\text{Pick-up}) = .4 + .2 = .6\)
Such calculations illustrate the additive nature of probabilities when considering mutually exclusive events, and they underscore the importance of accurately identifying all aspects of the sample space.
Probability Theory Basics
Probability theory is the mathematical framework that deals with calculating the likelihood of different events. It is based on a set of axioms, which together with an understanding of sample spaces and event probabilities, provide the tools needed to evaluate random phenomena across disciplines.

Basic principles include concepts like mutually exclusive events—events that cannot occur simultaneously. When events are mutually exclusive, like the delivery options in our example, the probability of any one of them occurring is the sum of their individual probabilities if there's no overlap.

The opposite would be non-exclusive events, where outcomes can occur in tandem. In such cases, probability calculations must adjust for the overlap. Probability theory also deals with independent events (one event’s occurrence doesn't affect another's) and dependent events (one event’s outcome influences another's possibility).

Using probability theory, we can elucidate the uncertainty that comes with various real-world scenarios, whether it's in choosing delivery options or predicting weather patterns, helping us make more informed decisions amidst uncertainty.

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Most popular questions from this chapter

Delayed diagnosis of cancer is a problem because it can delay the start of treatment. The paper "Causes of Physician Delay in the Diagnosis of Breast Cancer" (Archives of Internal Medicine \([2002]: 1343-1348)\) examined possible causes for delayed diagnosis for women with breast cancer. The accompanying table summarizes data on the initial written mammogram report (benign or suspicious) and whether or not diagnosis was delayed for 433 women with breast cancer. $$ \begin{array}{l|cc} & & \text { Diagnosis } \\ & \begin{array}{c} \text { Diagnosis } \\ \text { Delayed } \end{array} & \begin{array}{c} \text { Not } \\ \text { Delayed } \end{array} \\ \hline \begin{array}{l} \text { Mammogram Report Benign } \\ \text { Mammogram Report } \\ \text { Suspicious } \end{array} & 32 & 89 \\ & 8 & 304 \\ \hline \end{array} $$ Consider the following events: \(B=\) the event that the mammogram report says benign \(S=\) event that the mammogram report says suspicious \(D=\) event that diagnosis is delayed a. Assume that these data are representative of the larger group of all women with breast cancer. Use the data in the table to find and interpret the following probabilities: i. \(\quad P(B)\) ii. \(P(S)\) iii. \(P(D \mid B)\) iv. \(P(D \mid S)\) b. Remember that all of the 433 women in this study actually had breast cancer, so benign mammogram reports were, by definition, in error. Write a few sentences explaining whether this type of error in the reading of mammograms is related to delayed diagnosis of breast cancer.

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.

Consider the following information about travelers on vacation: \(40 \%\) check work e-mail, \(30 \%\) use a cell phone to stay connected to work, \(25 \%\) bring a laptop with them on vacation, \(23 \%\) both check work e-mail and use a cell phone to stay connected, and \(51 \%\) neither check work e-mail nor use a cell phone to stay connected nor bring a laptop. In addition \(88 \%\) of those who bring a laptop also check work e-mail and \(70 \%\) of those who use a cell phone to stay connected also bring a laptop. With \(E=\) event that a traveler on vacation checks work e-mail, \(C=\) event that a traveler on vacation uses a cell phone to stay connected, and \(L=\) event that a traveler on vacation brought a laptop, use the given information to determine the following probabilities. A Venn diagram may help. a. \(P(E)\) b. \(P(C)\) c. \(P(L)\) d. \(P(E\) and \(C)\) e. \(P\left(E^{C}\right.\) and \(C^{C}\) and \(\left.L^{C}\right)\) f. \(P(E\) and \(\operatorname{Cor} L)\) g. \(\quad P(E \mid L)\) h. \(P(L \mid C)\) i. \(P(E\) and \(C\) and \(L)\) j. \(P(E\) and \(L)\) k. \(P(C\) and \(L)\) 1\. \(P(C \mid E\) and \(L)\)

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