Question

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.
What is the probability that someone who tests positive does not have the genetic disease?
What is the probability that someone who tests negative has the disease?

Answer 1

Answer:

Part a: The probability that someone tests positive and does not have the disease is given as P(F'|E) is 0.6667

Part b: The probability that someone tests negative and  have the disease is given as P(F|E') is 0.000000103

Step-by-step explanation:

The event that the test is positive is  E and the patient has disease is  F

so

The probability of a person having disease is  P(F)=1/10000=0.0001

The probability of a person having disease and test positive is  P(E|F)=0.999

The probability of a person not having disease and testing positive is  P(E|F')=0.0002

Now by using complement rule, the value of people not having the disease is  P(F')=1-P(F)=1-0.0001=0.9999

Part a:

The probability that someone tests positive and does not have the disease is given as P(F'|E) by Bayes theorem is as

$P(F'|E)=\dfrac{P(E|F')P(F')}{P(E|F)P(F)+P(E|F')P(F')}$

By substitution of the values

$P(F'|E)=\dfrac{P(E|F')P(F')}{P(E|F)P(F)+P(E|F')P(F')}\\P(F'|E)=\dfrac{(0.0002)(0.9999)}{(0.9999)(0.0001)+(0.0002)(0.9999)}\\P(F'|E)=\dfrac{0.00019998}{0.00029997}\\P(F'|E)=0.6667$

The probability that someone tests positive and does not have the disease is given as P(F'|E) is 0.6667

Part b

The probability that someone tests positive and does not have the disease is given as P(F|E') by Bayes theorem is as

$P(F|E')=\dfrac{P(E'|F)P(F)}{P(E'|F)P(F)+P(E'|F')P(F')}$

The probabilities are calculated as

P(E'|F)=1-P(E|F)=1-0.999=0.001

P(E'|F')=1-P(E|F')=1-0.0002=0.9998

By substitution of the values

$P(F|E')=\dfrac{P(E'|F)P(F)}{P(E'|F)P(F)+P(E'|F')P(F')}\\P(F|E')=\dfrac{(0.001)(0.0001)}{(0.001)(0.0001)+(0.9998)(0.9999)}\\P(F|E')=\dfrac{0.0000001}{0.99970012}\\P(F|E')=0.000000103$

The probability that someone tests negative and have the disease is given as P(F|E') is 0.000000103