Question

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 a 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 is chosen and rows corresponding to whether or not gift wrap is selected. b. Use the table to calculate \(P(E \cup G)\). Give a long-run relative frequency interpretation of this probability.

Short answer

The probability of a customer choosing expedited shipping or gift wrap, or both, denoted by \(P(E \cup G)\) is approximately 0.3568 or 35.68%. This means that if we were to simulate this scenario with 1000 customers many times, on average about 357 customers would choose either expedited shipping or gift wrap or both in each simulation.

Step-by-step solution

Understanding the Problem

It's understood that customers can either choose or not to choose expedited shipping and gift wrapping. These two events are termed as 'E' and 'G' respectively. The problem also provides the probability for each event, and states these events are independent - meaning that the occurrence of one event does not affect the occurrence of the other.

Constructing a 'Hypothetical 1000' Table

Based on the given probabilities, the number of customers who will choose expedited shipping out of 1000 will be \(0.26 \times 1000 = 260\). Those who will choose gift wrapping will be \(0.12 \times 1000 = 120\). Because these two events are independent, the number of customers who will choose both expedited shipping and gift wrapping will be the product of their individual probabilities, which is \(0.26 \times 0.12 \times 1000 = 31.2\). The table will be like: \n\n\n| Choice | Expedited Shipping | No Expedited Shipping | TOTAL |\n| ------------- | ----------------: | --------------------: | ----: |\n| Gift wrap | 31.2 | 88.8 | 120 |\n| No Gift wrap | 228.8 | 651.2 | 880 |\n| TOTAL | 260 | 740 | 1000 |

Calculating \(P(E \cup G)\)

The union of two events E and G, denoted E U G, refers to either E, or G, or both happening at the same time. The probability of the union of two independent events is given by the sum of the probabilities of each event, minus the probability of the intersection of the two events. Therefore, \(P(E \cup G) = P(E) + P(G) - P(E)P(G) = 0.26 + 0.12 - 0.26 \times 0.12 = 0.3568\).

Providing Long-run Relative Frequency Interpretation

A long-run relative frequency interpretation of this probability means that in the long run, out of 1000 customers, approximately 357 will either choose expedited shipping or gift wrap or both.

Key concepts

Independent Events Probability

When we talk about independent events in probability, we are referring to the scenario where the outcome of one event does not influence the occurrence of another. In other words, knowing whether one event has happened does not change the likelihood of the other event happening.

For instance, in the exercise provided, the decision to choose expedited shipping (Event E) is independent from the decision to choose gift wrapping (Event G). This is an important concept in probability because it simplifies the calculation of the likelihood of both events occurring together. To determine the probability that both independent events happen, we simply multiply their individual probabilities:
\[ P(E \text{ and } G) = P(E) \cdot P(G) \].

In practical terms, if a customer's decision to expedite shipping has no impact on their choice to have their items gift wrapped, each choice is made without regard to the other. This characteristic of independence is pivotal to understanding how different probabilities can interact within a statistical context.

Hypothetical 1000 Table

A 'hypothetical 1000' table is a powerful illustration tool used in statistics to envision the outcomes of events based on their probabilities. By scaling up to 1000 cases, it becomes easier to conceptualize the probabilities as tangible counts, aiding in understanding and calculation.

In this exercise, constructing the table involved multiplying the probability of each event by 1000. This provides an estimate of how many times we would expect each event to occur if we repeated the situation 1000 times. The table is divided into categories—such as with or without expedited shipping and with or without gift wrapping—and the numbers filled in each cell reflect the expected frequency. This visual representation simplifies the calculation of joint and union probabilities and makes the abstract concept of probability more concrete.

For example, to determine the number of customers who might choose both expedited shipping and gift wrap, we use the independent probabilities of E and G. By the independence property, we multiply these together and then by 1000, which yields an estimate that serves to populate the corresponding cell of our table.

Long-run Relative Frequency Interpretation

The long-run relative frequency interpretation of probability is the idea that if an experiment were repeated many times, the proportion of a particular outcome would approach a certain value. This is the 'long-run' probability of the event. It is a practical approach, especially when discussing independent events, because it considers the outcomes of multiple trials.

In the given problem, when we talk about the probability of customers choosing either expedited shipping or gift wrapping or both, we calculate the long-run relative frequency as the measure of likelihood. The calculation of \(P(E \cup G)\) gives us a way to predict and interpret the outcomes over an infinite number of trials or customers, though we often illustrate it with a large but finite number like 1000. So, the long-run relative frequency tells us that if we observe 1000 customers, approximately 357 of them will either choose expedited shipping or gift wrapping or both. This interpretation is helpful because it transforms abstract probability into real-world expectations that can inform decisions and predictions.