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Chapter 6: Problem 31

The article "Chances Are You Know Someone with a Tattoo, and He's Not a Sailor" (Associated Press, June 11, 2006) included results from a survey of adults aged 18 to 50 . The accompanying data are consistent with summary values given in the article. $$ \begin{array}{l|cc} & \text { At Least One Tattoo } & \text { No Tattoo } \\ \hline \text { Age 18-29 } & 18 & 32 \\ \text { Age 30-50 } & 6 & 44 \\ \hline \end{array} $$ Assuming these data are representative of adult Americans and that an adult American is selected at random, use the given information to estimate the following probabilities. a. \(P(\) tattoo \()\) b. \(P(\) tattoo \(\mid\) age \(18-29)\) c. \(P(\) tattoo \(\mid\) age \(30-50\) ) d. \(P(\) age \(18-29 \mid\) tattoo \()\)

Short Answer

Expert verified
a. \(P(\text{tattoo}) = 0.24\), b. \(P(\text{tattoo | age 18-29}) = 0.36\), c. \(P(\text{tattoo | age 30-50}) = 0.12\), d. \(P(\text{age 18-29 | tattoo}) = 0.75\)

Step by step solution

01

Calculate Total

First, we calculate the total number of people surveyed by summing up all the numbers in our table. In the table, there are 18 + 32 people aged 18-29 and 6 + 44 people aged 30-50 which gives us a total of 100 people.
02

Calculate P(Tattoo)

To find the total probability of having a tattoo, we add the number of people who have tattoos (18 from age group 18-29 and 6 from age group 30-50 to get 24). We then divide this by the total number of people to get a probability of \(\frac{24}{100}=0.24\).
03

Calculate P(Tattoo|Age 18-29)

To find the probability of having a tattoo given the person is between 18-29 years, we consider only people in this age group. The number of people aged between 18-29 with a tattoo is 18, and the total number of people aged between 18-29 is 18 + 32 = 50. Therefore, the probability is \(\frac{18}{50}=0.36\).
04

Calculate P(Tattoo|Age 30-50)

To find the probability of having a tattoo given the person is between 30-50 years, we consider only people in this age group. The number of people aged between 30-50 with a tattoo is 6, and the total number of people aged between 30-50 is 6 + 44 = 50. Therefore, the probability is \(\frac{6}{50}=0.12\).
05

Calculate P(Age 18-29|Tattoo)

To find the probability of being aged between 18-29 given that the person has a tattoo, we consider only people who have a tattoo. The number of people with a tattoo between 18-29 years is 18, and the total number of people with a tattoo is 18 + 6 = 24. Therefore, the probability is \(\frac{18}{24}=0.75\).

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Key Concepts

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

Statistical Probability
Statistical probability is the measure of the likelihood that an event will occur based on statistical data and historical evidence. It is essentially the ratio of the favorable outcomes to the total number of possible outcomes, often expressed as a percentage or a number between 0 and 1. In the context of our exercise, we used a table that presented data on the number of adults with and without tattoos across two different age groups. To estimate the statistical probability of a randomly selected adult having a tattoo, we calculated the total number of people with tattoos and divided by the overall population considered in the survey.

This approach is widely used in various fields to make informed decisions or predictions based on past occurrences. It's essential to note that the accuracy of statistical probability highly depends on the quality and representativeness of the data. If the survey data in our problem truly represents the adult American population, then the calculated probability of \(P(tattoo)\) is a reliable estimate.
Conditional Probability
Conditional probability is a measure of the probability of an event occurring given that another event has already occurred. It helps us to refine our predictions based on new information. The notation \(P(A | B)\) is used to denote the probability of event A occurring given that B has already happened.

In our example, when we calculated \(P(\text{tattoo} | \text{age } 18-29)\), we were looking for the probability of an individual having a tattoo on the condition that they belong to the 18-29 age group. Similarly, the calculation of \(P(\text{age } 18-29 | \text{tattoo})\) gives us the likelihood that a person is aged 18-29 given they have a tattoo. These conditional probabilities are valuable in understanding the demographics of tattoo ownership within given age groups and can be used by marketers, health professionals, and sociologists to tailor interventions or campaigns.
Probability Calculation
Probability calculation involves determining the chance of a particular event happening. It requires counting the number of times an event can occur and the total number of outcomes. The formula commonly used is \(P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\). One of the best practices for accurate probability calculation is to ensure that categories do not overlap and that all possible outcomes are accounted for.

In the exercise, we performed several probability calculations, each demonstrating a different principle of probability. When calculating the probability of someone having a tattoo in the general adult population, we used the basic probability formula. For conditional probabilities, we modified the denominator to reflect the condition, like focusing on a specific age group only. Each calculation provided different insights, highlighting the importance of knowing which probability type to calculate in a given scenario.

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Most popular questions from this chapter

A new model of laptop computer can be ordered with one of three screen sizes ( 10 inches, 12 inches, 15 inches) and one of four hard drive sizes \((50 \mathrm{~GB}, 100 \mathrm{~GB}\), \(150 \mathrm{~GB}\), and \(200 \mathrm{~GB}\) ). Consider the chance experiment in which a laptop order is selected and the screen size and hard drive size are recorded. a. Display possible outcomes using a tree diagram. b. Let \(A\) be the event that the order is for a laptop with a screen size of 12 inches or smaller. Let \(B\) be the event that the order is for a laptop with a hard drive size of at most \(100 \mathrm{~GB}\). What outcomes are in \(A^{C}\) ? In \(A \cup B ?\) In \(A \cap B\) ? c. Let \(C\) denote the event that the order is for a laptop with a \(200 \mathrm{~GB}\) hard drive. Are \(A\) and \(C\) disjoint events? Are \(B\) and \(C\) disjoint?

"N.Y. Lottery Numbers Come Up 9-1-1 on \(9 / 11 "\) was the headline of an article that appeared in the San Francisco Chronicle (September 13, 2002). More than 5600 people had selected the sequence \(9-1-1\) on that date, many more than is typical for that sequence. \(\mathrm{A}\) professor at the University of Buffalo is quoted as saying, "I'm a bit surprised, but I wouldn't characterize it as bizarre. It's randomness. Every number has the same chance of coming up.' a. The New York state lottery uses balls numbered \(0-9\) circulating in three separate bins. To select the winning sequence, one ball is chosen at random from each bin. What is the probability that the sequence \(9-1-1\) is the sequence selected on any particular day? (Hint: It may be helpful to think about the chosen sequence as a three-digit number.) b. What approach (classical, relative frequency, or subjective) did you use to obtain the probability in Part (a)? Explain.

Consider a Venn diagram picturing two events \(A\) and \(B\) that are not disjoint. a. Shade the event \((A \cup B)^{C} .\) On a separate Venn diagram shade the event \(A^{C} \cap B^{C} .\) How are these two events related? b. Shade the event \((A \cap B)^{C} .\) On a separate Venn diagram shade the event \(A^{C} \cup B^{C} .\) How are these two events related? (Note: These two relationships together are called DeMorgan's laws.)

An article in the New york Times (March 2. 1994) reported that people who suffer cardiac arrest in New York City have only a 1 in 100 chance of survival. Using probability notation, an equivalent statement would be \(P(\) survival \()=.01\) for people who suffer a cardiac arrest in New York City. (The article attributed this poor survival rate to factors common in large cities: traffic congestion and the difficulty of finding victims in large buildings.) a. Give a relative frequency interpretation of the given probability. b. The research that was the basis for the New York Times article was a study of 2329 consecutive cardiac arrests in New York City. To justify the " 1 in 100 chance of survival" statement, how many of the 2329 cardiac arrest sufferers do you think survived? Explain.

A college library has four copies of a certain book; the copies are numbered \(1,2,3\), and 4 . Two of these are selected at random. The first selected book is placed on 2-hour reserve, and the second book can be checked out overnight. a. Construct a tree diagram to display the 12 outcomes in the sample space. b. Let \(A\) denote the event that at least one of the books selected is an even-numbered copy. What outcomes are in \(A\) ? c. Suppose that copies 1 and 2 are first printings, whereas copies 3 and 4 are second printings. Let \(B\) denote the event that exactly one of the copies selected is a first printing. What outcomes are contained in \(B\) ?
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