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According to a study released by the federal Substance Abuse and Mental Health Services Administration (Knight Ridder Tribune, September 9,1999 ), approximately \(8 \%\) of all adult full-time workers are drug users and approximately \(70 \%\) of adult drug users are employed full-time. a. Is it possible for both of the reported percentages to be correct? Explain. b. Define the events \(D\) and \(E\) as \(D=\) event that a randomly selected adult is a drug user and \(E=\) event that a randomly selected adult is employed full- time. What are the estimated values of \(P(D \mid E)\) and \(P(E \mid D) ?\) c. Is it possible to determine \(P(D)\), the probability that a randomly selected adult is a drug user, from the information given? If not, what additional information would be needed?

Short Answer

Expert verified
a. The reported percentages cannot both be correct as they imply contradictory percentages of adult drug users. b. The estimated values are \(P(D \mid E) = 0.08\) and \(P(E \mid D) = 0.70\). c. It is not possible to determine \(P(D)\) without additional information.

Step by step solution

01

Assessing Simultaneity of Given Percentages

First, let's consider if the two given percentages can hold truth simultaneously. Necessarily, if \(8\%\) of all adult full-time workers are drug users, it means that more than \(8\%\) of all adults must be drug users, because not all adults work full-time. However, if \(70\%\) of all adult drug users are employed full-time, it suggests that less than \(70\%\) of all adults could be drug users, because not all adults are employed full-time. Because of these contradictory implications, it's not possible for both percentages to be correct simultaneously.
02

Calculation of \(P(D \mid E)\) and \(P(E \mid D)\)

Next, let's determine the values of \(P(D \mid E)\) and \(P(E \mid D)\). From the exercise, \(P(D \mid E)\) is the probability that a randomly selected adult who is employed full-time is a drug user, which is given as \(8\%\). Hence, \(P(D \mid E) = 0.08\). Similarly, \(P(E \mid D)\) is the probability that a randomly selected drug user is employed full-time, which is given as \(70\%\). Hence, \(P(E \mid D) = 0.70\).
03

Possibility of Determining \(P(D)\)

Finally, let's see if we can determine \(P(D)\), the probability that a randomly selected adult is a drug user. To calculate \(P(D)\), we would need additional information such as the total number of adults and the total number of drug users. Without this information, we cannot determine \(P(D)\).

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Key Concepts

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

Conditional Probability
Conditional probability is a fundamental concept in probability theory that deals with the likelihood of an event occurring given that another event has already occurred. This is denoted as \( P(A \mid B) \), which reads as 'the probability of A given B'. To understand conditional probability, imagine flipping a coin and then rolling a die. If you want to know the probability of rolling a 4 given that the coin landed on heads, you are considering a conditional probability.

In our exercise, conditional probability is used to determine \( P(D \mid E) \) and \( P(E \mid D) \). This signifies the probability of being a drug user given that the adult is employed full-time (\( P(D \mid E) \)) and vice versa (\( P(E \mid D) \)). It's important for students to recognize these as examples of conditional probabilities because they show how the probability of one event can be affected by the occurrence of another event.
Event Probability
Event probability refers to the measure of the chance that a particular event will occur. This is often represented as a fraction or a percentage ranging from 0 (the event will definitely not occur) to 1 (the event will definitely occur). In probability theory, any outcome or combination of outcomes from a random process is considered an 'event'.

When calculating the probability of an event, one must consider all the possible outcomes. For example, the probability of rolling a 3 on a six-sided die is \( \frac{1}{6} \) because there is one favourable outcome (rolling a 3) out of six possible outcomes. The exercise provided deals with an event probability when it asks us to determine the probability that a randomly selected adult is a drug user (\( P(D) \)), based on the given percentage of all adult full-time workers who are drug users. To calculate this, as mentioned in the solution, additional information about the total population is required.
Probability Theory
Probability theory is the branch of mathematics that studies random events and quantifies uncertainty. It provides a mathematical foundation for predicting the likelihood of various outcomes. The principles of probability theory apply to a wide range of activities, including analysis of games of chance, weather forecasting, and risk assessment in finance and insurance.

Probability theory works with concepts like event probability, conditional probability, and the addition and multiplication rules for probability, among others. In the context of the exercise, probability theory is applied to assess the relationship between adult full-time workers and drug users, determining if the percentages given can coexist logically and what data would be needed to confirm the probability of an adult being a drug user. Understanding the underlying principles of probability theory allows for a systematic approach to solving such real-world problems.

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Most popular questions from this chapter

Consider the chance experiment in which an automobile is selected and both the number of defective headlights \((0,1\), or 2\()\) and the number of defective tires \((0,1\), 2,3, or 4 ) are determined. a. Display possible outcomes using a tree diagram. b. Let \(A\) be the event that at most one headlight is defective and \(B\) be the event that at most one tire is defective. What outcomes are in \(A^{C} ?\) in \(A \cup B ?\) in \(A \cap B ?\) c. Let \(C\) denote the event that all four tires are defective. Are \(A\) and \(C\) disjoint events? Are \(B\) and \(C\) disjoint?

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Suppose that a box contains 25 bulbs, of which 20 are good and the other 5 are defective. Consider randoml selecting three bulbs without replacement. Let \(E\) denote the event that the first bulb selected is good, \(F\) be the event that the second bulb is good, and \(G\) represent the event that the third bulb selected is good. a. What is \(P(E)\) ? b. What is \(P(F \mid E)\) ? c. What is \(P(G \mid E \cap F)\) ? d. What is the probability that all three selected bulbs are good?

The National Public Radio show Car Talk has a feature called "The Puzzler." Listeners are asked to send in answers to some puzzling questions-usually about cars but sometimes about probability (which, of course, must account for the incredible popularity of the program!). Suppose that for a car question, 800 answers are submitted, of which 50 are correct. Suppose also that the hosts randomly select two answers from those submitted with replacement. a. Calculate the probability that both selected answers are correct. (For purposes of this problem, keep at least five digits to the right of the decimal.) b. Suppose now that the hosts select the answers at random but without replacement. Use conditional probability to evaluate the probability that both answers selected are correct. How does this probability compare to the one computed in Part (a)?

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