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Chapter 6: Problem 74

Only \(0.1 \%\) of the individuals in a certain population have a particular disease (an incidence rate of .001). Of those who have the disease, \(95 \%\) test positive when 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 \(\mid\) positive test). Does the result surprise you? Give an intuitive explanation for the size of this probability.

Short Answer

Expert verified
The probabilities are: a) For individuals having the disease and testing positive, it is \(0.00095\); b) The overall probability of testing positive is \(0.10085\); c) The probability of having the disease given a positive test is \(0.00942\). The result may be surprisingly low due to the high incidence of false positive tests arising from low disease prevalence.

Step by step solution

01

Construct a tree diagram

A. Construct a tree diagram with first-generation branches representing having the disease and not having the disease. Under having disease, attach two second-generation branches indicating testing positive and testing negative. Do likewise for the 'not having a disease' branch. The probabilities that go with each branch are: has disease \(0.1\%\), doesn't have disease \(99.9\%\), for those with disease, test positive \(95\%\) and test negative \(5\%\), for those without disease, test positive \(10\%\) and test negative \(90\%\).
02

Calculation of \(P(\) has disease and positive test).

B. Apply the general multiplication rule, which is \(P(A and B) = P(A)鈭桺(B|A)\), where A signifies 'has disease' and B denotes 'tests positive'. For those having the disease, the probability of testing positive is \(0.001 * 0.95 = 0.00095\).
03

Calculation of \(P\) (positive test).

C. The probability of a positive test is the sum of the probabilities of those who have the disease and test positive, plus those who do not have the disease but also test positive. That is \(P(\)Disease Positive) + P(\)No Disease Positive) = \(0.00095 + 0.999 * 0.1 = 0.10085\).
04

Calculation of \(P\) (has disease | positive test.

D. The probability of having the disease given a positive test is calculated by applying Bayes' theorem: \(P(\)Disease|Positive) = \(\frac{P(Disease Positive)}{P(Positive)}\) = \(\frac{0.00095}{0.10085} = 0.00942\). This probability may seem unexpectedly low. The reason is that even though the test is quite accurate, the disease is very rare. So, most people who test positive are actually healthy people who received a false positive test result.

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Key Concepts

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

Tree Diagram
A tree diagram is a useful tool in probability to visually represent and organize different possible outcomes of a series of events. It's particularly helpful when dealing with complex probability problems, like those involving more than one stage or step.
To construct a tree diagram, you begin by identifying the main outcomes of the initial event, which are represented as the first-generation branches. For example, in the given exercise, the population is divided into two groups: those who have the disease and those who don鈥檛. These make up the first set of branches.
Each of these branches then splits into further branches, representing possible outcomes of the next event. In this case, the second-generation branches are "tests positive" or "tests negative".
  • If a person has the disease, there are branches for testing positive and testing negative.
  • If a person does not have the disease, there are also branches for testing positive and testing negative.
On each branch, you place the probability of that outcome, which helps in analyzing and calculating the overall probabilities subsequently.
Multiplication Rule
The multiplication rule in probability is essential when calculating the likelihood of two independent events occurring together. It can be stated as:
\[P(A \text{ and } B) = P(A) \times P(B|A)\]This means you multiply the probability of the first event, \(P(A)\), by the conditional probability of the second event occurring given that the first event has already occurred, \(P(B|A)\).
This rule was applied in our exercise to calculate the probability of having the disease and testing positive. In this context, "event A" is someone having the disease, which has a probability of 0.1%, or 0.001. "Event B" is testing positive, which has a probability of 95%, or 0.95, given the person has the disease.
Thus, the probability of both having the disease and testing positive is the product of these probabilities:
\[0.001 \times 0.95 = 0.00095\]This straightforward application of the multiplication rule allows us to evaluate scenarios with sequential events efficiently.
Incidence Rate
The incidence rate is a measure used in probability and epidemiology to describe how common a particular event or condition is in a specified population over a certain time period. It is usually expressed as a percentage or a proportion.
In the context of the exercise, the incidence rate of the disease in the given population is 0.1%, or 0.001. This rate reflects the proportion of individuals within the population who have the disease.
Understanding the incidence rate is critical because it significantly influences the probability calculations and interpretations. Even if a test is highly accurate, as is shown here with the disease's test, a low incidence rate can lead to unexpected results such as a small probability of actually having the disease given a positive test result. This highlights why context and base rates are crucial considerations in interpreting diagnostic test results and other probability problems.
Bayes' Theorem
Bayes' theorem is a fundamental formula in the field of probability that helps update the probability estimation for an event based on new evidence. It is expressed as:
\[P(A|B) = \frac{P(B|A) \times P(A)}{P(B)}\]Where \(P(A|B)\) is the probability of event A occurring given event B has occurred, \(P(B|A)\) is the probability of event B given A, \(P(A)\) is the probability of A, and \(P(B)\) is the probability of B.
In this exercise, Bayes' theorem is used to find the probability of having the disease given a positive test result. This involves calculating:
  • \(P(\text{Disease|Positive}) = \frac{P(\text{Positive|Disease}) \times P(\text{Disease})}{P(\text{Positive})}\)
Here, the probability \(P(\text{Positive})\) previously calculated is 0.10085, the probability \(P(\text{Positive|Disease})\) is 0.95, and \(P(\text{Disease})\) is 0.001. Plugging these in yields:
\[P(\text{Disease|Positive}) = \frac{0.95 \times 0.001}{0.10085} \approx 0.00942\]This result, although seeming counterintuitive due to its small size, is explained by considering that the disease's rarity and the likelihood of false positives impact the overall probability drastically, thus demonstrating the power and necessity of Bayesian probability for nuanced decision-making.

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Most popular questions from this chapter

A family consisting of three people- \(\mathrm{P}_{1}, \mathrm{P}_{2}\), and \(\mathrm{P}_{3}\) \- belongs to a medical clinic that always has a physician at each of stations 1,2, and 3 . During a certain week, each member of the family visits the clinic exactly once and is randomly assigned to a station. One experimental outcome is \((1,2,1)\), which means that \(\mathrm{P}_{1}\) is assigned to station \(1 .\) \(\mathrm{P}_{2}\) to station 2, and \(\mathrm{P}_{3}\) to station 1 a. List the 27 possible outcomes. (Hint: First list the nine outcomes in which \(\mathrm{P}_{1}\) goes to station 1 , then the nine in which \(\mathrm{P}_{1}\) goes to station 2 , and finally the nine in which \(\mathrm{P}_{1}\) goes to station 3 ; a tree diagram might help.) b. List all outcomes in the event \(A\), that all three people go to the same station. c. List all outcomes in the event \(B\), that all three people go to different stations. d. List all outcomes in the event \(C\), that no one goes to station \(2 .\) e. Identify outcomes in each of the following events: \(B^{C}\), \(C^{C}, A \cup B, A \cap B, A \cap C\)

Suppose that a new Internet company Mumble.com requires all employees to take a drug test. Mumble.com can afford only the inexpensive drug test - the one with a \(5 \%\) false-positive rate and a \(10 \%\) false-negative rate. (That means that \(5 \%\) of those who are not using drugs will incorrectly test positive and that \(10 \%\) of those who are actually using drugs will test negative.) Suppose that \(10 \%\) of those who work for Mumble.com are using the drugs for which Mumble is checking. (Hint: It may be helpful to draw a tree diagram to answer the questions that follow.) a. If one employee is chosen at random, what is the probability that the employee both uses drugs and tests positive? b. If one employee is chosen at random, what is the probability that the employee does not use drugs but tests positive anyway? c. If one employee is chosen at random, what is the probability that the employee tests positive? d. If we know that a randomly chosen employee has tested positive, what is the probability that he or she uses drugs?

A transmitter is sending a message using a binary code, namely, a sequence of 0's and 1's. Each transmitted bit \((0\) or 1\()\) must pass through three relays to reach the receiver. At each relay, the probability is \(.20\) that the bit sent on is different from the bit received (a reversal). Assume that the relays operate independently of one another: transmitter \(\rightarrow\) relay \(1 \rightarrow\) relay \(2 \rightarrow\) relay \(3 \rightarrow\) receiver a. If a 1 is sent from the transmitter, what is the probability that a 1 is sent on by all three relays? b. If a 1 is sent from the transmitter, what is the probability that a 1 is received by the receiver? (Hint: The eight experimental outcomes can be displayed on a tree diagram with three generations of branches, one generation for each relay.)

Two different airlines have a flight from Los Angeles to New York that departs each weekday morning at a certain time. Let \(E\) denote the event that the first airline's flight is fully booked on a particular day, and let \(F\) denote the event that the second airline's flight is fully booked on that same day. Suppose that \(P(E)=.7, P(F)=.6\), and \(P(E \cap F)=.54\). a. Calculate \(P(E \mid F)\) the probability that the first airline's flight is fully booked given that the second airline's flight is fully booked. b. Calculate \(P(F \mid E)\).

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 \(\mathrm{A}\), 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 \(P(\) 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.
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