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Chapter 6: Problem 73

In an article that appears on the web site of the American Statistical Association (www.amstat.org), Carlton Gunn, a public defender in Seattle, Washington, wrote about how he uses statistics in his work as an attorney. He states: I personally have used statistics in trying to challenge the reliability of drug testing results. Suppose the chance of a mistake in the taking and processing of a urine sample for a drug test is just 1 in 100 . And your client has a "dirty" (i.e., positive) test result. Only a 1 in 100 chance that it could be wrong? Not necessarily. If the vast majority of all tests given- say 99 in \(100-\) are truly clean, then you get one false dirty and one true dirty in every 100 tests, so that half of the dirty tests are false. Define the following events as \(T D=\) event that the test result is dirty, \(T C=\) event that the test result is clean, \(D=\) event that the person tested is actually dirty, and \(C=\) event that the person tested is actually clean. a. Using the information in the quote, what are the values of \mathbf{i} . ~ \(P(T D \mid D)\) iii. \(P(C)\) ii. \(P(T D \mid C)\) iv. \(P(D)\) b. Use the law of total probability to find \(P(T D)\). c. Use Bayes' rule to evaluate \(P(C \mid T D)\). Is this value consistent with the argument given in the quote? Explain.

Short Answer

Expert verified
The respective probabilities are \(P(T D \mid D) = 1\), \(P(T D \mid C) = 0.01\), \(P(D) = 0.01\) and \(P(C) = 0.99\). The probability that the test is dirty \(P(T D) = 0.02\). Also, the probability that a person is clean given that the test result is dirty \(P(C \mid T D)= 0.495\). These results are consistent with the assertion made in the quote.

Step by step solution

01

Identifying and Calculating Probabilities

The given probabilities from the problem can be defined as follows: \\(P(T D \mid D)\) is the probability that the test result is dirty given the person is actually dirty. This is a true positive and its value is 1 because if a person is actually dirty, the test is correct. \\(P(T D \mid C)\) is the probability that the test result is dirty given the person is actually clean. This is a false positive, which indicates errors in testing and its value is 1/100 or 0.01 because there's 1% error. \\(P(D)\) is the probability that a person is actually dirty. This value is 1/100 or 0.01 because one in every 100 tests is true dirty. \\(P(C)\) is the probability that a person is actually clean which is 99/100 or 0.99 since 99 in every 100 tests are truly clean.
02

Using the Law of Total Probability to Find \(P(T D)\)

The law of total probability states that the probability of an event can be found by considering all the ways that it could happen, summing products of conditional probability and the probability of each condition. Here, \(P(T D) = P(T D \mid D) * P(D) + P(T D \mid C) * P(C)\). Substituting the values, we get \(P(T D) = 1 * 0.01 + 0.01 * 0.99 = 0.02\)
03

Using Bayes' Rule to Evaluate \(P(C \mid T D)\)

Bayes' theorem allows us to revise predictions in light of new data. It is given by \(P(C \mid T D) = \frac{P(T D \mid C) * P(C)}{P(T D)}\). Upon substituting the values we get \(P(C \mid T D) = \frac{0.01 * 0.99}{0.02} = 0.495\). This indicates that there is approximately 50% chance that a person is clean given that the test result is dirty. This is consistent with what was argued in the quote in the question where it was mentioned that 'half of the dirty tests are false'.

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Key Concepts

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

Conditional Probability
Conditional probability provides a way to find the probability of an event occurring given that another event has already occurred. In our context, let's say you want to find the probability of the test result being dirty (T D) given certain conditions about the person's actual status - whether they are truly dirty (D) or clean (C). This draws directly from the concept of conditional probability.

You may encounter such terms as:- \(P(T D \mid D)\), which shows the probability of a dirty result if the test subject is truly dirty.- \(P(T D \mid C)\), which is the probability of a dirty result if the test subject is actually clean.

Understanding these definitions helps you interpret drug test results more accurately. It highlights how errors can occur even with a seemingly small probability of false positives. These calculations use conditional probability to give clarity on the reliability of test outcomes.

Law of Total Probability
The Law of Total Probability is a fundamental rule that connects marginal probabilities to conditional probabilities. It helps compute the likelihood of an event by considering all possible underlying scenarios, each weighted by their respective probabilities.

In this exercise, you see it in action while calculating the probability of the test showing dirty (T D). With the probabilities \(P(T D \mid D)\) and \(P(T D \mid C)\) already known, the law aids by summing these conditional probabilities, each multiplied by the probability of each condition (clean or dirty).

Here's how it works: \[P(T D) = P(T D \mid D) \times P(D) + P(T D \mid C) \times P(C)\] where - \(P(T D \mid D)\) and \(P(T D \mid C)\) are conditional probabilities we've defined earlier.- \(P(D)\) and \(P(C)\) represent prior probabilities of being dirty or clean.

This equation effectively combines information about various sources of a "dirty" result, yielding a total probability that represents real-world testing conditions more fully.

Bayes' Theorem
Bayes' Theorem is a powerful tool in statistics that allows us to update our beliefs based on new evidence. It's particularly useful when you want to calculate the probability of a certain hypothesis, given some observed data.

From the exercise, it demonstrates how you can find the probability that a person is actually clean (C) given a dirty test result (T D). This formula you use in this scenario looks like:\[P(C \mid T D) = \frac{P(T D \mid C) \times P(C)}{P(T D)}\]

Where:- \(P(T D \mid C)\) is the probability a test shows dirty when in fact the person is clean.- \(P(C)\) is the probability that a person is clean overall.- \(P(T D)\) is the total probability of receiving a dirty test.

By applying Bayes' Theorem, you find that around 49.5%, or nearly half the time, a positive test result could be an error for someone actually clean. This insight matches the scenario described in Gunn's argument from the quote, showing how crucial Bayes鈥� Theorem is for decision-making in uncertain conditions.

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Most popular questions from this chapter

The manager of a music store has kept records of the number of CDs bought in a single transaction by customers who make a purchase at the store. The accompanying table gives six possible outcomes and the estimated probability associated with each of these outcomes for the chance experiment that consists of observing the number of CDs purchased by the next customer at the store. $$ \begin{aligned} &\begin{array}{l} \text { Number of CDs } \\ \text { purchased } \end{array} & 1 & 2 & 3 & 4 & 5 & 6 \text { or more } \\ &\begin{array}{c} \text { Estimated } \\ \text { probability } \end{array} & .45 & .25 & .10 & .10 & .07 & .03 \end{aligned} $$ a. What is the estimated probability that the next customer purchases three or fewer CDs? b. What is the estimated probability that the next customer purchases at most three CDs? How does this compare to the probability computed in Part (a)? c. What is the estimated probability that the next customer purchases five or more CDs? d. What is the estimated probability that the next customer purchases one or two CDs? e. What is the estimated probability that the next customer purchases more than two CDs? Show two different ways to compute this probability that use the probability rules of this section.

Only \(0.1 \%\) of the individuals in a certain population have a particular disease (an incidence rate of .001). Of th贸se wh貌 have the disease, \(95 \%\) test possitive wh猫n a certain diagnostic test is applied. Of those who do not have the disease, \(90 \%\) test negative when the test is applied. Suppose that an individual from this population is randomly selected and given the test. a. Construct a tree diagram having two first-generation branches, for has disease and doesn't have disease, and two second-generation branches leading out from each of these, for positive test and negative test. Then enter appropriate probabilities on the four branches. b. Use the general multiplication rule to calculate \(P\) (has disease and positive test). c. Calculate \(P\) (positive test). d. Calculate \(P\) (has disease| positive test). Does the result surprise you? Give an intuitive explanation for the size of this probability.

Suppose that a six-sided die is "loaded" so that any particular even-numbered face is twice as likely to land face up as any particular odd-numbered face. Consider the chance experiment that consists of rolling this die. a. What are the probabilities of the six simple events? (Hint: Denote these events by \(O_{1}, \ldots, O_{6}\). Then \(P\left(O_{1}\right)=p, P\left(O_{2}\right)=2 p, P\left(O_{3}\right)=p, \ldots, P\left(O_{6}\right)=2 p\) Now use a condition on the sum of these probabilities to determine \(p\).) b. What is the probability that the number showing is an odd number? at most three? c. Now suppose that the die is loaded so that the probability of any particular simple event is proportional to the number showing on the corresponding upturned face; that is, \(P\left(O_{1}\right)=c, P\left(O_{2}\right)=2 c, \ldots\), \(P\left(O_{6}\right)=6 c\). What are the probabilities of the six simple events? Calculate the probabilities of Part (b) for this die.

A company uses three different assembly lines \(A_{1}, A_{2}\), and \(A_{3}\) -to manufacture a particular component. Of those manufactured by \(A_{1}, 5 \%\) need rework to remedy a defect, whereas \(8 \%\) of \(A_{2}\) 's components and \(10 \%\) of \(A_{3}\) 's components need rework. Suppose that \(50 \%\) of all components are produced by \(A_{1}\), whereas \(30 \%\) are produced by \(A_{2}\) and \(20 \%\) come from \(A_{3}\). a. Construct a tree diagram with first-generation branches corresponding to the three lines. Leading from each branch, draw one branch for rework (R) and another for no rework (N). Then enter appropriate probabilities on the branches. b. What is the probability that a randomly selected component came from \(A_{1}\) and needed rework? c. What is the probability that a randomly selected component needed rework?

After all students have left the classroom, a statistics professor notices that four copies of the text were left under desks. At the beginning of the next lecture, the professor distributes the four books at random to the four students \((1,2,3\), and 4\()\) who claim to have left books. One possible outcome is that 1 receives 2's book, 2 receives 4's book, 3 receives his or her own book, and 4 receives l's book. This outcome can be abbreviated \((2,4,3,1)\). a. List the 23 other possible outcomes. b. Which outcomes are contained in the event that exactly two of the books are returned to their correct owners? Assuming equally likely outcomes, what is the probability of this event? c. What is the probability that exactly one of the four students receives his or her own book? d. What is the probability that exactly three receive their own books? e. What is the probability that at least two of the four students receive their own books?
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