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Chapter 5: Problem 92

An online store offers two methods of shipping-regular ground service and an expedited 2 -day shipping. Customers may also choose whether or not to have the purchase gift wrapped. Suppose that the events \(E=\) event that the customer chooses expedited shipping \(G=\) event that the customer chooses gift wrap are independent with \(P(E)=0.26\) and \(P(G)=0.12\). a. Construct a hypothetical 1000 table with columns corresponding to whether or not expedited shipping was chosen and rows corresponding to whether or not gift wrap was selected. b. Use the information in the table to calculate \(P(E \cup G)\). Give a long- run relative frequency interpretation of this probability.

Short Answer

Expert verified
In this problem, we found that the probabilities of a customer choosing expedited shipping, gift wrapping, or both are independent. We calculated the probability of each combination of the events and used it to construct a hypothetical 1000-customer table. We then calculated the probability of the union of the events, P(E 鈭� G), to be 0.63. This means that in the long run, approximately 63% of customers of the online store choose either expedited shipping, gift wrapping, or both when making a purchase.

Step by step solution

01

Determine the probability of each combination of shipping method and gift wrap choice

Since events E and G are independent, we can calculate the probability of each combination of shipping method and gift wrap choice using the given probabilities of events E and G. There are four combinations: 1. Expedited Shipping and Gift Wrap (E 鈭� G): P(E 鈭� G) = P(E) 路 P(G) = 0.26 路 0.12 = 0.0312 2. Expedited Shipping and No Gift Wrap (E 鈭� G'): P(E 鈭� G') = P(E) 路 P(G') = 0.26 路 (1 - 0.12) = 0.26 路 0.88 = 0.2288 3. Regular Shipping and Gift Wrap (E' 鈭� G): P(E' 鈭� G) = P(E') 路 P(G) = (1 - 0.26) 路 0.12 = 0.74 路 0.12 = 0.0888 4. Regular Shipping and No Gift Wrap (E' 鈭� G'): P(E' 鈭� G') = P(E') 路 P(G') = 0.74 路 0.88 = 0.6512
02

Construct the hypothetical 1000-customer table

Using the probabilities calculated in step 1, we can now construct the table with 1000 customers. To find the number of customers for each combination of the shipping method and gift wrap choice, multiply the probabilities by 1000: 1. Expedited Shipping and Gift Wrap (E 鈭� G): 312 customers 2. Expedited Shipping and No Gift Wrap (E 鈭� G'): 228.8 鈮� 229 customers 3. Regular Shipping and Gift Wrap (E' 鈭� G): 88.8 鈮� 89 customers 4. Regular Shipping and No Gift Wrap (E' 鈭� G'): 651.2 鈮� 651 customers
03

Calculate P(E 鈭� G) using the table

The probability of E 鈭� G can be calculated from the table as follows: P(E 鈭� G) = P(E 鈭� G) + P(E 鈭� G') + P(E' 鈭� G) = (312 + 229 + 89) / 1000 = 630 / 1000 = 0.63
04

Give a long-run relative frequency interpretation of P(E 鈭� G)

The long-run relative frequency interpretation of P(E 鈭� G) = 0.63 means that in the long run, approximately 63% of the customers of the online store choose either expedited shipping, gift wrapping, or both when making a purchase.

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Key Concepts

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

Understanding Independent Events
When considering the interaction between different choices or occurrences, we introduce the concept of independent events. In the realm of probability, two events are independent if the occurrence of one event does not influence the occurrence of the other.

For instance, in our textbook exercise, the customer's decision to select expedited shipping is separate from their decision to choose gift wrapping. Knowing that a customer chose expedited shipping doesn't change the likelihood they'll opt for gift-wrapping, since these decisions are made independently of each other.

This characteristic allows us to perform specific probability calculations, such as multiplying the probability of one event by the probability of another to find the joint probability of the two occurring together, which is exactly what we see with the probabilities of expedited shipping (\(P(E) = 0.26\)) and opting for gift wrap (\(P(G) = 0.12\)).
Joint Probability: Merging Paths
The joint probability refers to the likelihood of two events occurring simultaneously. To compute this, we multiply the probabilities of the two independent events. In our online shipping example, the joint probability of customers selecting both expedited shipping and gift wrapping, denoted as \(P(E \cap G)\), can be calculated by multiplying \(P(E)\) and \(P(G)\).

With given probabilities \(P(E) = 0.26\) and \(P(G) = 0.12\), we find \(P(E \cap G) = 0.26 \times 0.12 = 0.0312\), indicating a 3.12% chance of both expedited shipping and gift wrapping being selected by any given customer. This allows us to predict outcomes and make informed decisions about resource allocation and marketing strategies.
Navigating Probability Calculations
Conducting probability calculations often involves a series of steps to determine the likelihood of various combinations of events. After establishing that events are independent, as we have with event E (expedited shipping) and event G (gift wrapping), we use basic arithmetic operations to find the probability for each possible combination of event occurrences.

For example, to calculate the probability that a customer chooses either expedited shipping and/or gift wrapping, we add the probabilities of E and G occurring separately and then also include their joint probability. This is reflected in the solution process, where the calculations are used to understand customer behaviors at a granular level, informing the store's operations and providing a better customer experience.
Long-run Relative Frequency Interpretation
The long-run relative frequency interpretation of a probability is a practical way of understanding likelihood over a large number of trials or instances. It suggests that if we were to repeat an experiment infinitely, the relative frequency with which a certain event occurs would tend towards its probability.

In our problem, the probability of a customer choosing expedited shipping, gift wrapping, or both was found to be \(P(E \cup G) = 0.63\) or 63%. This implies that, over a large number of transactions, we would expect approximately 63 out of every 100 customers to select at least one of the services. It's a helpful concept that resonates well with repeated processes or services, like those of an online store, providing a realistic expectation of customer choices over time.

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Most popular questions from this chapter

The report "Twitter in Higher Education: Usage Habits and Trends of Today's College Faculty" describes a survey of nearly 2000 college faculty. The report indicates the following: \(30.7 \%\) reported that they use Twitter, and \(69.3 \%\) said that they do not use Twitter. Of those who use Twitter, \(39.9 \%\) said they sometimes use Twitter to communicate with students. Of those who use Twitter, \(27.5 \%\) said that they sometimes use Twitter as a learning tool in the classroom. Consider the chance experiment that selects one of the study participants at random. a. Two of the percentages given in the problem specify unconditional probabilities, and the other two percentages specify conditional probabilities. Which are conditional probabilities? How can you tell? b. Suppose the following events are defined: \(T=\) event that selected faculty member uses Twitter \(C=\) event that selected faculty member sometimes uses Twitter to communicate with students \(L=\) event that selected faculty member sometimes uses Twitter as a learning tool in the classroom Use the given information to determine the following probabilities: i. \(P(T)\) iii. \(P(C \mid T)\) ii. \(P\left(T^{C}\right)\) iv. \(\quad P(L \mid T)\) c. Construct a hypothetical 1000 table using the given probabilities and use the information in the table to calculate \(P(C),\) the probability that the selected study participant sometimes uses Twitter to communicate with students. d. Construct a hypothetical 1000 table using the given probabilities and use the information in the table to calculate the probability that the selected study participant sometimes uses Twitter as a learning tool in the classroom.

a. Suppose events \(E\) and \(F\) are mutually exclusive with \(P(E)=0.64\) and \(P(F)=0.17\) i. What is the value of \(P(E \cap F)\) ? ii. What is the value of \(P(E \cup F)\) ? b. Suppose that \(A\) and \(B\) are events with \(P(A)=0.3, P(B)=0.5\), and \(P(A \cap B)=0.15 .\) Are \(A\) and \(B\) mutually exclusive? How can you tell? c. Suppose that \(A\) and \(B\) are events with \(P(A)=0.65\) and \(P(B)=0.57 .\) Are \(A\) and \(B\) mutually exclusive? How can you tell?

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