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Chapter 4: Problem 111

Two thousand randomly selected adults were asked whether or not they have ever shopped on the Internet. The following table gives a two-way classification of the responses. $$\begin{array}{lcc} & \text { Have Shopped } & \text { Have Never Shopped } \\ \hline \text { Male } & 500 & 700 \\ \text { Female } & 300 & 500 \\ \hline \end{array}$$ Suppose one adult is selected at random from these 2000 adults. Find the following probabilities. a. \(P(\) has never shopped on the Internet or is a female) b. \(P(\) is a male \(o r\) has shopped on the Internet) c. \(P\) (has shopped on the Internet or has never shopped on the Internet)

Short Answer

Expert verified
a: 0.75, b: 0.5, c: 1

Step by step solution

01

- Calculate Total Responses

First, sum up all the responses. This gives the total number of events, which is 2000 (500 males have shopped + 700 males never shopped + 300 females have shopped + 500 females never shopped).
02

- Compute for Part (a)

To find the probability that the adult selected at random has never shopped on the Internet or is a female, sum the total number of adults who have never shopped and the total number of females, and then divide by the total number of adults. This results to \[P(\) has never shopped on the Internet or is a female) = (700 male never shopped + 500 female never shopped + 300 female have shopped) / 2000 = 1500 / 2000 = 0.75\].
03

- Compute for Part (b)

To find the probability that the adult selected is a male or has shopped on the Internet, sum the total number of males and the total number of adults who have shopped, and then subtract the intersection (males who have shopped) to avoid double counting, then divide by the total number of adults. This results to \[P(\) is a male \(o r\) has shopped on the Internet) = (500 males have shopped + 700 male never shopped + 500 female have shopped + 300 female have shopped - 500 males have shopped) / 2000 = 1000 / 2000 = 0.5\].
04

- Compute for Part (c)

The probability that an adult has shopped on the Internet or has never shopped on the Internet actually covers all possible outcomes. This is because every adult in the sample either has or has not shopped on the Internet, no other options are possible. So, \[P(\) has shopped on the Internet or has never shopped on the Internet) = 1\].

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Key Concepts

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

Two-way classification
A two-way classification is a method used to organize data into a table. It allows us to categorize and analyze results based on two different factors or characteristics. In the given exercise, the two factors involved are: gender (male or female) and shopping experience (have shopped or have never shopped online).
This classification provides a clear visual representation of data, making it easier to analyze relationships between the factors. For instance, we can quickly see how many males have shopped online or how many females have never done so.
  • Two-way tables help in displaying categorical data.
  • They aid in visualizing the comparison between different groups.

By understanding two-way classifications, it is easier to compute probabilities based on these categories, which leads us to our next concept: random selection.
Random selection
Random selection refers to a method where each individual in a group has an equal chance of being chosen. In probability, this approach is crucial to avoid bias. Imagine picking a name out of a hat, where every slip of paper is folded the same way; this ensures fairness.
In the exercise, one adult is selected at random from a group of 2,000. This means each adult had the same probability (1 out of 2,000) of being selected.
  • Ensures fairness, eliminating preferential treatment.
  • Gives equal probability to all individuals within the defined group.

Random selection is fundamental in probability because it provides a reliable basis from which accurate probability calculations arise. Next, let's delve into what this means for the sample space.
Sample space
In probability, the sample space is the entire set of possible outcomes of an experiment. It's all the potential results that could occur. For the exercise, the sample space consists of all 2,000 adults surveyed.
The sample space can also be broken down into more specific categories based on the two-way classification. For example, the sample space includes:
  • 500 males who have shopped online
  • 700 males who have never shopped online
  • 300 females who have shopped online
  • 500 females who have never shopped online

Understanding the sample space is vital as it lays the groundwork for probability computations. Knowing the total number of possible outcomes allows us to determine the likelihood of a particular event.
Probability computation
Probability computation involves calculating the likelihood of a specific event happening. It is done by dividing the number of successful outcomes by the total number of possible outcomes in the sample space.
For instance, to find the probability of selecting someone who has never shopped online or is a female from the exercise, we use the formula:
\[P(\text{{has never shopped or is a female}}) = \frac{{700 + 500 + 300}}{2000} = 0.75\]
Here鈥檚 why this works:
  • "Successful outcomes" refer to the combined totals of all adults fitting the criteria (i.e., have never shopped or are female).
  • The probability is a ratio of this sum to the total number of adults.
Probability computations allow for informed predictions based on quantitative data analysis. Understanding how to perform these computations empowers students to solve probability-based problems effectively.

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Most popular questions from this chapter

A trimotor plane has three engines-a central engine and an engine on each wing. The plane will crash only if the central engine fails and at least one of the two wing engines fails. The probability of failure during any given flight is \(.005\) for the central engine and \(.008\) for each of the wing engines. Assuming that the three engines operate independently, what is the probability that the plane will crash during a flight?

According to a March 2009 Gallup Poll (http://www.gallup.com/poll/1 17025/Support-NuclearEnergy-Inches-New-High.aspx), \(71 \%\) of Republicans/Republican leaners and \(52 \%\) of Democrats/Democrat leaners favor the use of nuclear power. The survey consisted of 1012 American adults, approximately half of whom were Republicans or Republican leaners. Suppose the following table gives the distribution of responses of these 1012 adults $$\begin{array}{lcc} \hline & \text { Favor } & \text { Do not favor } \\ \hline \text { Republicans/Republican leaners } & 381 & 128 \\ \text { Democrats/Democrat leaners } & 258 & 245 \\ \hline \end{array}$$ a. If one person is selected at random from this sample of 1012 U.S. adults, find the probability that this person i. does not favor the use of nuclear power ii. is a Republican/Republican leaner iii. favors the use of nuclear power given that the person is a Republican/Republican leaner jv. is a Republican/Republican leaner given that the person does not favor the use of nuclear power b. Are the events favors and does not favor mutually exclusive? What about the events does not favor and Republican/Republican leaner? c. Are the events does not favor and Republican/Republican leaner independent? Why or why not?

Given that \(A, B\), and \(C\) are three independent events, find their joint probability for the following. a. \(P(A)=.20, \quad P(B)=.46\), and \(P(C)=.25\) b. \(P(A)=.44, \quad P(B)=.27\), and \(P(C)=.43\)

The probability is \(.80\) that a senior from a large college in New York State has never gone to Florida for spring break. If two college seniors are selected at random from this college, what is the probability that the first has never gone to Florida for spring break and the second has? Draw a tree diagram for this problem.

A random sample of 250 adults was taken, and they were asked whether they prefer watching sports or opera on television. The following table gives the two-way classification of these adults $$\begin{array}{lcc} \hline & \begin{array}{c} \text { Prefer Watching } \\ \text { Sports } \end{array} & \begin{array}{c} \text { Prefer Watching } \\ \text { Opera } \end{array} \\ \hline \text { Male } & 96 & 24 \\ \text { Female } & 45 & 85 \end{array}$$ a. If one adult is selected at random from this group, find the probability that this adult i. prefers watching opera ii. is a male iii. prefers watching sports given that the adult is a female iv. is a male given that he prefers watching sports \(\mathbf{v}\). is a female and prefers watching opera vi. prefers watching sports or is a male b. Are the events "female" and "prefers watching sports" independent? Are they mutually exclusive? Explain why or why not.
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