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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

Two individuals, \(A\) and \(B\), are finalists for a chess championship. They will play a sequence of games, each of which can result in a win for \(A\), a win for \(\mathrm{B}\), or a draw. Suppose that the outcomes of successive games are independent, with \(P(\) A wins game \()=.3\), \(P(\) B wins game \()=.2\), and \(P(\) draw \()=.5\). Each time a player wins a game, he earns 1 point and his opponent earns no points. The first player to win 5 points wins the championship. For the sake of simplicity, assume that the championship will end in a draw if both players obtain 5 points at the same time. a. What is the probability that \(A\) wins the championship in just five games? b. What is the probability that it takes just five games to obtain a champion? c. If a draw earns a half-point for each player, describe how you would perform a simulation to estimate \(\mathrm{P}(\mathrm{A}\) wins the championship). d. If neither player earns any points from a draw, would the simulation in Part (c) take longer to perform? Explain your reasoning.

Many fire stations handle emergency calls for medical assistance as well as calls requesting firefighting equipment. A particular station says that the probability that an incoming call is for medical assistance is .85. This can be expressed as \(P(\) call is for medical assistance \()=.85\). a. Give a relative frequency interpretation of the given probability. b. What is the probability that a call is not for medical assistance? c. Assuming that successive calls are independent of one another, calculate the probability that two successive calls will both be for medical assistance. d. Still assuming independence, calculate the probability that for two successive calls, the first is for medical assistance and the second is not for medical assistance. e. Still assuming independence, calculate the probability that exactly one of the next two calls will be for medical assistance. (Hint: There are two different possibilities. The one call for medical assistance might be the first call, or it might be the second call.) f. Do you think that it is reasonable to assume that. the requests made in successive calls are independent? Explain.

Is ultrasound a reliable method for determining the gender of an unborn baby? The accompanying data on 1000 births are consistent with summary values that appeared in the online version of the Journal of Statistics Education ("New Approaches to Leaming Probability in the First Statistics Course" [2001]). $$ \begin{array}{ccc} & \begin{array}{c} \text { Ultrasound } \\ \text { Predicted } \\ \text { Female } \end{array} & \begin{array}{c} \text { Ultrasound } \\ \text { Predicted } \\ \text { Male } \end{array} \\ \hline \begin{array}{c} \text { Actual Gender Is } \\ \text { Female } \end{array} & 432 & 48 \\ \begin{array}{c} \text { Actual Gender Is } \\ \text { Male } \end{array} & 130 & 390 \\ \hline \end{array} $$ a. Use the given information to estimate the probability that a newborn baby is female, given that the ultrasound predicted the baby would be female. b. Use the given information to estimate the probability that a newborn baby is male, given that the ultrasound predicted the baby would be male. c. Based on your answers to Parts (a) and (b), do you think that a prediction that a baby is male and a prediction that a baby is female are equally reliable? Explain.

A bookstore sells two types of books (fiction and nonfiction) in several formats (hardcover, paperback, digital, and audio). For the chance experiment that consists of observing the type and format of a single-book purchase, two possible outcomes are a hardcover fiction book and an audio nonfiction book. a. There are eight outcomes in the sample space for this experiment. List these possible outcomes. b. Do you think it is reasonable to think that the outcomes for this experiment would be equally likely? Explain. c. For customers who purchase a single book, the estimated probabilities for the different possible outcomes are given in the cells of the accompanying table. What is the probability that a randomly selected single-book purchase will be for a book in print format (hardcover or paperback)? $$ \begin{array}{l|cccc} {\text { Hardcover }} & \text { Paperback } & \text { Digital } & \text { Audio } \\ \hline \text { Fiction } & .15 & .45 & .10 & .10 \\ \text { Nonfiction } & .08 & .04 & .02 & .06 \\ \hline \end{array} $$ d. Show two different ways to compute the probability that a randomly selected single-book purchase will be for a book that is not in a print format. e. Find the probability that a randomly selected singlebook purchase will be for a work of fiction.

A shipment of 5000 printed circuit boards contains 40 that are defective. Two boards will be chosen at random, without replacement. Consider the two events \(E_{1}=\) event that the first board selected is defective and \(E_{2}=\) event that the second board selected is defective. a. Are \(E_{1}\) and \(E_{2}\) dependent events? Explain in words. b. Let \(n o t E_{1}\) be the event that the first board selected is not defective (the event \(E_{1}^{C}\) ). What is \(P\left(\right.\) not \(E_{1}\) )? c. How do the two probabilities \(P\left(E_{2} \mid E_{1}\right)\) and \(P\left(E_{2} \mid\right.\) not \(\left.E_{1}\right)\) compare? d. Based on your answer to Part (c), would it be reasonable to view \(E_{1}\) and \(E_{2}\) as approximately independent?
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