/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 A genetic test is used to determ... [FREE SOLUTION] | 91影视

91影视

A genetic test is used to determine if people have a predisposition for thrombosis, which is the formation of a blood clot inside a blood vessel that obstructs the flow of blood through the circulatory system. It is believed that \(3 \%\) of people actually have this predisposition. The genetic test is \(99 \%\) accurate if a person actually has the predisposition, meaning that the probability of a positive test result when a person actually has the predisposition is \(0.99 .\) The test is \(98 \%\) accurate if a person does not have the predisposition. What is the probability that a randomly selected person who tests positive for the predisposition by the test actually has the predisposition?

Short Answer

Expert verified
Probability is approximately 60.59%.

Step by step solution

01

Understand the Problem

We need to determine the probability that a person has a predisposition for thrombosis given that they tested positive. This can be calculated using Bayes' theorem, a way to find a conditional probability.
02

Identify and Define Probabilities

Let:- \( P(D) \) be the probability a person has the predisposition \( (0.03) \).- \( P(T^+|D) \) be the probability of testing positive given the person has the predisposition \( (0.99) \).- \( P(T^-|eg D) \) be the probability of testing negative given the person does not have the predisposition \( (0.98) \).- \( P(eg D) \) be the probability a person does not have the predisposition \( (0.97) \).- \( P(T^+|eg D) \) be the probability of testing positive given the person does not have the predisposition, hence \( 1 - P(T^-|eg D) = 0.02 \).
03

Use Bayes' Theorem

Bayes' theorem formula is: \[ P(D|T^+) = \frac{P(T^+|D)P(D)}{P(T^+)} \].We need to first calculate \( P(T^+) \), the probability of testing positive, using law of total probability: \[ P(T^+) = P(T^+|D)P(D) + P(T^+|eg D)P(eg D) \].
04

Calculate P(T鈦)

Plug in the known probabilities into the formula: \[ P(T^+) = (0.99)(0.03) + (0.02)(0.97) \].Calculate the values:- \( P(T^+|D)P(D) = 0.0297 \)- \( P(T^+|eg D)P(eg D) = 0.0194 \).Add them: \[ P(T^+) = 0.0297 + 0.0194 = 0.0491 \].
05

Calculate P(D|T鈦)

Now substitute into Bayes' theorem:\[ P(D|T^+) = \frac{0.0297}{0.0491} \].Perform the division to find:\[ P(D|T^+) \approx 0.6059 \].
06

Interpret the Result

This result means that if a person tests positive for thrombosis predisposition, there is approximately a 60.59% chance they actually have the predisposition.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Conditional Probability
Conditional probability is a core concept in probability theory that allows us to find the likelihood of an event occurring, given that another event has already occurred. It essentially refines the probability based on the context provided by the known event.
When considering a genetic test for conditions such as thrombosis predisposition, conditional probability helps in calculating how likely it is that a person has the predisposition if they've tested positive.

To find such probabilities, we use the notation \( P(A|B) \), which represents the probability of event \( A \) occurring given that \( B \) is true.
  • For instance, in the exercise above, \( P(D|T^+) \) represents the probability of having the predisposition \( D \) given that the test was positive \( T^+ \).
  • This builds upon the initial likelihood \( P(D) \) of any person having the predisposition, which was given as \( 3\% \).
Understanding conditional probability is crucial for correctly interpreting test results, especially in medical testing contexts.
Genetic Testing
Genetic testing is a medical tool used to detect genetic predispositions to certain diseases or conditions, like thrombosis.
These tests analyze specific genes or genetic markers to inform individuals about potential health risks.

In the exercise, the genetic test for thrombosis predisposition is both a practical application of probability and a real-world example of genetic testing鈥檚 potential.
  • The test is \( 99\% \) accurate if someone has the predisposition, which means it correctly identifies those individuals nearly all the time (true positives).
  • Conversely, the test is also \( 98\% \) accurate when someone doesn't have the predisposition, implying it mostly correctly identifies those without the predisposition (true negatives).
However, tests can yield false results. Understanding how these results translate into probabilities can help us make informed decisions based on test outcomes.
Law of Total Probability
The law of total probability is a fundamental rule in probability theory that helps calculate the total probability of an event based on different scenarios.
It is particularly useful when considering events that can occur through various mutually exclusive paths.

This law is used in the exercise to determine the overall probability of testing positive for thrombosis predisposition, \( P(T^+) \).
  • By considering both scenarios鈥攚hether a person has the predisposition or not鈥攖he law allows us to find the total likelihood of getting a positive result.
  • It splits the probability into two paths: one where the person has the trait \( D \), and one where they do not \(eg D \).
The probability of testing positive \( P(T^+) \) is calculated as: \[ P(T^+) = P(T^+|D)P(D) + P(T^+|eg D)P(eg D) \]This ensures all potential ways of testing positive are considered and informs subsequent calculations.
Probability Calculation
Probability calculation is essential for solving real-world problems involving uncertainty, like determining health risks.
It enables us to bring mathematical precision to situations where complete certainty is impossible.

In practice, probability calculation involves steps like identifying relevant probabilities and applying mathematical formulas or theorems.
  • For this case, we begin by identifying the individual probabilities: \( P(D) \), \( P(T^+|D) \), \( P(eg D) \), and \( P(T^+|eg D) \).
  • We then employ Bayes' Theorem, a formula used to update the probability estimate for an event based on additional information, to find \( P(D|T^+) \).
The resulting formula gives us \[ P(D|T^+) = \frac{P(T^+|D)P(D)}{P(T^+)} \],allowing a more accurate prediction regarding medical concerns, ensuring results are interpreted correctly.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A Pew Research poll asked 1,306 Americans "From what you've read and heard, is there solid evidence that the average temperature on earth has been getting warmer over the past few decades, or not?". The table below shows the distribution of responses by party and ideology, where the counts have been replaced with relative frequencies. $$\begin{array}{lcccc} &{\text { Response }} & \\ & \begin{array}{c}\text { Earth is } \\ \text { warming }\end{array} & \begin{array}{c}\text { Not } \\\\\text { warming } \end{array} & \begin{array}{c}\text { Don't Know } \\\\\text { Refuse }\end{array} & \text { Total } \\ \hline \text { Conservative Republican } & 0.11 & 0.20 & 0.02 & 0.33 \\ \text { Mod/Lib Republican } & 0.06 & 0.06 & 0.01 & 0.13 \\ \text { Mod/Cons Democrat } & 0.25 & 0.07 & 0.02 & 0.34 \\ \text { Liberal Democrat } & 0.18 & 0.01 & 0.01 & 0.20 \\ \hline \text { Total } & 0.60 & 0.34 & 0.06 & 1.00 \end{array}$$ (a) Are believing that the earth is warming and being a liberal Democrat mutually exclusive? (b) What is the probability that a randomly chosen respondent believes the earth is warming or is a liberal Democrat? (c) What is the probability that a randomly chosen respondent believes the earth is warming given that he is a liberal Democrat? (d) What is the probability that a randomly chosen respondent believes the earth is warming given that he is a conservative Republican? (e) Does it appear that whether or not a respondent believes the earth is warming is independent of their party and ideology? Explain your reasoning. (f) What is the probability that a randomly chosen respondent is a moderate/liberal Republican given that he does not believe that the earth is warming?

Sally gets a cup of coffee and a muffin every day for breakfast from one of the many coffee shops in her neighborhood. She picks a coffee shop each morning at random and independently of previous days. The average price of a cup of coffee is $$\$ 1.40$$ with a standard deviation of $$30 \mathrm{c}(\$ 0.30),$$ the average price of a muffin is $$\$ 2.50$$ with a standard deviation of \(15 \mathrm{c},\) and the two prices are independent of each other. (a) What is the mean and standard deviation of the amount she spends on breakfast daily? (b) What is the mean and standard deviation of the amount she spends on breakfast weekly ( 7 days)?

Guessing on an exam. In a multiple choice exam, there are 5 questions and 4 choices for each question \((\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}) .\) Nancy has not studied for the exam at all and decides to randomly guess the answers. What is the probability that: (a) the first question she gets right is the \(5^{t h}\) question? (b) she gets all of the questions right? (c) she gets at least one question right?

Data collected at elementary schools in DeKalb County, GA suggest that each year roughly \(25 \%\) of students miss exactly one day of school, \(15 \%\) miss 2 days, and \(28 \%\) miss 3 or more days due to sickness. \(^{24}\) (a) What is the probability that a student chosen at random doesn't miss any days of school due to sickness this year? (b) What is the probability that a student chosen at random misses no more than one day? (c) What is the probability that a student chosen at random misses at least one day? (d) If a parent has two kids at a DeKalb County elementary school, what is the probability that neither kid will miss any school? Note any assumption you must make to answer this question. (e) If a parent has two kids at a DeKalb County elementary school, what is the probability that both kids will miss some school, i.e. at least one day? Note any assumption you make. (f) If you made an assumption in part (d) or (e), do you think it was reasonable? If you didn't make any assumptions, double check your earlier answers.

The Behavioral Risk Factor Surveillance System (BRFSS) is an annual telephone survey designed to identify risk factors in the adult population and report emerging health trends. The following table displays the distribution of health status of respondents to this survey (excellent, very good, good, fair, poor) and whether or not they have health insurance. $$\begin{array}{rrrrrrr} & {\text { Health Status }} & \\ & \text { Excellent } & \text { Very good } & \text { Good } & \text { Fair } & \text { Poor } & \text { Total } \\ \hline \text { No } & 0.0230 & 0.0364 & 0.0427 & 0.0192 & 0.0050 & 0.1262 \\ \text { Yes } & 0.2099 & 0.3123 & 0.2410 & 0.0817 & 0.0289 & 0.8738 \\ \hline \text { Total } & 0.2329 & 0.3486 & 0.2838 & 0.1009 & 0.0338 & 1.0000 \end{array}$$ (a) Are being in excellent health and having health coverage mutually exclusive? (b) What is the probability that a randomly chosen individual has excellent health? (c) What is the probability that a randomly chosen individual has excellent health given that he has health coverage? (d) What is the probability that a randomly chosen individual has excellent health given that he doesn't have health coverage? (e) Do having excellent health and having health coverage appear to be independent?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.