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91Ó°ÊÓ

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. $$ \begin{array}{lcc} \hline & \text { Have Shopped } & \text { Have Never Shopped } \\ \hline \text { Male } & 500 & 700 \\ \text { Female } & 300 & 500 \\ \hline \end{array} $$ 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
The probability for situation (a) is 0.65, for (b) it is 0.75 and for (c) it is 1.

Step by step solution

01

Identifying Total Outcomes

The first step would be to identify the total outcomes. In this case, we know that 2000 adults were randomly selected. So, our total number of outcomes is 2000.
02

Solve for Part (a)

Part (a) asks for the probability that an adult either has never shopped on the Internet or is a female. To find this, we'll add the number of adults who have never shopped on the Internet and the number of females, then substract the females who have never shopped on the Internet, because they are getting counted twice when summing. After that, divide by the total number of adults. In numbers: \((700 + 500 + 300 - 300) ÷ 2000 = 0.65\)
03

Solve for Part (b)

Part (b) asks for the probability that an adult is either a male or has shopped on the Internet. We apply the same technique as in step 2. So we take the amount of males, add the amount of adults who have shopped and substract the males who have shopped because they were counted twice. Then we divide this by the total amount of adults: \((1200 + 800 - 500) ÷ 2000 = 0.75\)
04

Solve for Part (c)

Part (c) asks for the probability that an adult has either shopped on the Internet, or has never shopped on the internet. As these are the only two options, all adults fall into this category. Therefore, the probability is \(1\).

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Key Concepts

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

Two-way Classification
In probability and statistics, a two-way classification table is a handy tool.
It helps categorize data into two different dimensions, allowing us to analyze relationships between categories. In the context of the original problem, we have two categories: gender (male or female) and shopping behavior (have shopped or have never shopped).

Imagine the table with rows and columns:
  • Rows represent genders (male and female).
  • Columns indicate whether they have shopped on the Internet or not.

This classification allows us to see the distribution and relationship between these groups quickly, providing the necessary data to calculate probabilities for different scenarios. The overlaps or intersections in the table, such as males who have shopped, are crucial for solving probability problems.
Understanding how these classifications connect will help you tackle a variety of probability questions with confidence.
Random Selection
Random selection is a fundamental concept in statistics. When we select something randomly, each entity has an equal chance of being chosen.

In the given problem, it refers to choosing one adult from a group of 2000 without bias. This ensures that every adult has the same probability of being selected, which is essential for fair and unbiased probability calculations.
Recognizing random selection helps us understand that our calculations are based on a principle of equality. Random samples allow for the derivation of probabilities that reflect the true characteristics of the entire group. This means any findings or probabilities we calculate are representative and not skewed by a systematic preference towards any subgroup within the population.
Event Probability
Probability measures how likely an event is to occur. In our exercise, we calculate the probability of three events.

  • Event a: An adult has never shopped or is female.
  • Event b: An adult is male or has shopped online.
  • Event c: An adult has either shopped or never shopped on the Internet.

To find an event's probability, we divide the number of favorable outcomes by the total number of possible outcomes. The formula looks like this: \[P( ext{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\] For example, in Event A, we calculated the probability that someone has never shopped on the Internet or is female. By using the two-way classification table, we identified overlaps and applied the correct adjustments to avoid double-counting. Mastering these steps will enable you to approach varied probability problems with organized and clear logic.
Complementary Probability
Complementary probability is a concept focused on the likelihood that a particular event does not happen.

When considering two complementary events, such as an event and its opposite, their probabilities will sum up to 1. This is because these events cover all possible outcomes.
Take Event C in our example: "has shopped on the Internet or has never shopped on the Internet." All adults either fall into one category or the other, with no other possibilities. Therefore, the probability of Event C is 1.
Mathematically, this can be expressed as: \[P( ext{Event}) + P( ext{Not Event}) = 1\]
Understanding complementary probability helps to confirm your calculations and provides another viewpoint in understanding how probabilities work in sorrounding environments. It ensures a comprehensive grasp of how events interplay and exhaust all possibilities.

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Most popular questions from this chapter

A certain state's auto license plates have three letters of the alphabet followed by a three-digit number. a. How many different license plates are possible if all three-letter sequences are permitted and any number from 000 to 999 is allowed? b. Arnold witnessed a hit-and-run accident. He knows that the first letter on the license plate of the offender's car was a B, that the second letter was an \(\mathrm{O}\) or a \(\mathrm{Q}\), and that the last number was a 5\. How many of this state's license plates fit this description?

Consider the following games with two dice. a. A gambler is going to roll a die four times. If he rolls at least one 6, you must pay him $$\$ 5 .$$ If he fails to roll a 6 in four tries, he will pay you $$\$ 5 .$$ Find the probability that you must pay the gambler. Assume that there is no cheating. b. The same gambler offers to let you roll a pair of dice 24 times. If you roll at least one double 6 , he will pay you $$\$ 10$$. If you fail to roll a double 6 in 24 tries, you will pay him $$\$ 10$$. The gambler says that you have a better chance of winning because your probability of success on each of the 24 rolls is $$1 / 36$$ and you have 24 chances. Thus, he says, your probability of winning $$\$ 10$$ is \(24(1 / 36)=2 / 3 .\) Do you agree with this analysis? If so, indicate why. If not, point out the fallacy in his argument, and then find the correct probability that you will win.

In how many ways can a sample (without replacement) of 5 items be selected from a population of 15 items?

Given that \(A, B\), and \(C\) are three independent events, find their joint probability for the following. a. \(P(A)=.81, \quad P(B)=.49\), and \(P(C)=.36\) b. \(P(A)=.02, \quad P(B)=.03\), and \(P(C)=.05\)

What is meant by the joint probability of two or more events? Give one example.

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