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Chapter 7: Problem 8

Suppose that one person in \(10,000\) people has a rare genetic disease. There is an excellent test for the disease; 99.9\(\%\) of people with the disease test positive and only 0.02\(\%\) who do not have the disease test positive. a) What is the probability that someone who tests positive has the genetic disease? b) What is the probability that someone who tests negative does not have the disease?

Short Answer

Expert verified
a) The probability is approximately 0.333 or 33.3%. b) The probability is approximately 0.9999 or 99.99%.

Step by step solution

01

- Define Event Probabilities

Define the events: Let A be the event that a person has the disease.Let B be the event that a person tests positive for the disease.P(A) = 1/10000 = 0.0001P(A') = 1 - P(A) = 0.9999P(B|A) = 0.999 (probability of testing positive if diseased)P(B|A') = 0.0002 (probability of testing positive if not diseased)
02

- Apply Bayes鈥� Theorem for Part (a)

Use Bayes' Theorem to find P(A|B), the probability of having the disease given a positive test result:\[ P(A|B) = \frac{P(B|A) \, P(A)}{P(B|A) \, P(A) + P(B|A') \, P(A')} \]Substitute the known values into the formula:\[ P(A|B) = \frac{0.999 \, (0.0001)}{0.999 \, (0.0001) + 0.0002 \, (0.9999)} \]
03

- Calculate Part (a)

Simplify the expression:\[ P(A|B) = \frac{0.0000999}{0.0000999 + 0.00019998} = \frac{0.0000999}{0.00029988} \]Compute the division:\[ P(A|B) \approx 0.333 \]
04

- Apply Conditional Probability for Part (b)

Use the law of total probability to find the probability of testing negative:Define C as the event that a person tests negative. Since C is the complement of event B, we have:\[ P(C) = 1 - P(B) \]Calculate P(B) first using total probability:\[ P(B) = P(B|A)P(A) + P(B|A')P(A') \]Substitute known values:\[ P(B) = (0.999)(0.0001) + (0.0002)(0.9999) = 0.0000999 + 0.00019998 = 0.00029988 \]Then find P(C):\[ P(C) = 1 - 0.00029988 = 0.99970012 \]
05

- Calculate Part (b)

Using conditional probability to find P(A鈥檤C), the probability of not having the disease given a negative test result:\[ P(A'|C) = \frac{P(C|A') \, P(A')}{P(C)} \]Given that P(C|A) = 1 - P(B|A) = 0.001 (probability of negative test if diseased), and P(C|A') = 1 - P(B|A') = 0.9998 (probability of negative test if not diseased), the simplified formula is:\[ P(A'|C) = \frac{0.9998 \, 0.9999}{0.99970012} \]Simplify and calculate:\[ P(A'|C) = \frac{0.99970012}{0.99970012} = 1 \] Thus, P(A'|C) \approx 0.9999

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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 the likelihood of an event occurring given that another event has already occurred. It is denoted by \( P(A|B) \), which represents the probability of event A happening given that B has happened.In this exercise, we use conditional probability to determine \( P(A|B) \): the probability of a person having the disease given that they tested positive. To calculate it, we use Bayes' Theorem.Bayes' Theorem is given by:\[ P(A|B) = \frac{P(B|A) \, P(A)}{P(B)} \]Here:
  • \( P(A) \) is the prior probability of having the disease.
  • \( P(B|A) \) is the probability of testing positive if one has the disease.
By substituting the given probabilities, we explore how these events influence each other under the scenario of genetic testing.
Genetic Testing
Genetic testing is a type of medical test that identifies changes in chromosomes, genes, or proteins to confirm or rule out a genetic condition. It plays a vital role in this exercise.For our scenario:
  • The probability of a person having the genetic disease is 0.0001 (or 1 in 10,000).
  • The probability of testing positive if diseased is very high at 99.9% (\( P(B|A) = 0.999 \)).
  • The probability of testing positive if not diseased is very low at 0.02% (\( P(B|A') = 0.0002 \)).
Computing probabilities like these, we can provide critical insights about the likelihood of having or not having a disease based on test results, which is crucial for making informed healthcare decisions.
Law of Total Probability
The law of total probability helps us find the probability of an event by considering all possible ways that event can occur. It is integral to calculating \( P(B) \) in Bayes' Theorem.For this problem:
  • We need \( P(B) \), the probability of testing positive, which we find using the formula: \[ P(B) = P(B|A)P(A) + P(B|A')P(A') \]
  • Substituting known values: \[ P(B) = (0.999)(0.0001) + (0.0002)(0.9999) = 0.0000999 + 0.00019998 = 0.00029988 \]
Understanding the law of total probability enables us to consider both scenarios of having the disease and not having the disease when calculating \( P(B) \). This comprehensive approach ensures we account for all outcomes to provide accurate probabilities.

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