Respuesta :
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
[tex]P(F'|E)=\dfrac{P(E|F')P(F')}{P(E|F)P(F)+P(E|F')P(F')}[/tex]
By substitution of the values
[tex]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[/tex]
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
[tex]P(F|E')=\dfrac{P(E'|F)P(F)}{P(E'|F)P(F)+P(E'|F')P(F')}[/tex]
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
[tex]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[/tex]
The probability that someone tests negative and have the disease is given as P(F|E') is 0.000000103
Using conditional probability, it is found that there is a:
- 0.6669 = 66.69% probability that someone who tests positive does not have the genetic disease.
- [tex]1 \times 10^{-7}[/tex] probability that someone who tests negative has the disease.
Conditional Probability
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
- P(B|A) is the probability of event B happening, given that A happened.
- [tex]P(A \cap B)[/tex] is the probability of both A and B happening.
- P(A) is the probability of A happening.
Question 1:
- Event A: Positive test.
- Event B: Does not have the disease.
The probabilities associated with a positive test are:
- 99.9% of 1 in 10000 = 0.0001(have the disease).
- 0.02% of 0.9999(does not have the disease).
Hence
[tex]P(A) = 0.999(0.0001) + 0.0002(0.9999) = 0.00029988[/tex]
The probability of both a positive test and not having the disease is:
[tex]P(A \cap B) = 0.0002(0.9999)[/tex]
The conditional probability is:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0002(0.9999)}{0.00029988} = 0.6669[/tex]
0.6669 = 66.69% probability that someone who tests positive does not have the genetic disease.
Question 2:
- Event A: Negative test.
- Event B: Has the disease.
The probabilities associated with a negative test are:
- 0.1% of 1 in 10000 = 0.0001(have the disease).
- 99.98% of 0.9999(does not have the disease).
Hence
[tex]P(A) = 0.001(0.0001) + 0.9998(0.9999) = 0.99970012[/tex]
The probability of both a negative test and having the disease is:
[tex]P(A \cap B) = 0.001(0.0001)[/tex]
The conditional probability is:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.001(0.0001)}{0.99970012} = 1 \times 10^{-7}[/tex]
[tex]1 \times 10^{-7}[/tex] probability that someone who tests negative has the disease.
A similar problem is given at https://brainly.com/question/14398287
