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Chapter 4: Problem 27

A random sample of 2000 adults showed that 1320 of them have shopped at least once on the Internet. What is the (approximate) probability that a randomly selected adult has shopped on the Internet?

Short Answer

Expert verified
The approximate probability that a randomly selected adult has shopped on the Internet is \(\frac{1320}{2000}\) or 66%.

Step by step solution

01

Identify the Total Sample Size

The first step is to identify the total size of the sampling group. From the problem, we know that 2000 adults were surveyed.
02

Identify the Number of Specified Outcomes

The problem states that 1320 out of the 2000 adults surveyed have shopped on the internet at least once. This is the number of 'success' outcomes we are interested in.
03

Calculate the Probability

To calculate the probability of an adult having shopped on the internet, divide the number of adults who have shopped on the internet (1320) by the total number of adults surveyed (2000). This can be represented as \(\frac{1320}{2000}\).
04

Simplify the Probability

Further simplify the calculated probability by reducing the fraction to its simplest form or converting it to a decimal or percentage. This would help make the probability easier to understand.

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Most popular questions from this chapter

Given that \(A\) and \(B\) are two mutually exclusive events, find \(P(A\) or \(B\) ) for the following. a. \(P(A)=.25\) and \(P(B)=.27\) b. \(P(A)=.58\) and \(P(B)=.09\)

The probability that an open-heart operation is successful is .84. What is the probability that in two randomly selected open-heart operations at least one will be successful? Draw a tree diagram for this experiment.

Briefly explain the difference between the marginal and conditional probabilities of events. Give one example of each.

In a sample survey, 1800 senior citizens were asked whether or not they have ever been victimized by a dishonest telemarketer. The following table gives the responses by age group. $$\begin{array}{l|llcc} & & & \begin{array}{c} \text { Have Been } \\ \text { Victimized } \end{array} & \begin{array}{c} \text { Have Never } \\ \text { Been Victimized } \end{array} \\ \hline & 60-69 & \text { (A) } & 106 & 698 \\ \text { Age } & 70-79 & \text { (B) } & 145 & 447 \\ & 80 \text { or over (C) } & 61 & 343 \\ \hline \end{array}$$ Suppose one person is randomly selected from these senior citizens. Find the following probabilities a. \(P(\) have been victimized or \(\mathrm{B}\) ) b. \(P(\) have never been victimized or \(\mathrm{C}\) )

Given that \(P(A)=.30\) and \(P(A\) and \(B)=.24\), find \(P(B \mid A)\).
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