Answer:
A)The probability that someone who tests positive has the disease is 0.9995
B)The probability that someone who tests negative does not have the disease is 0.99999
Step-by-step explanation:
Let D be the event that a person has a disease
Let
be the event that a person don't have a disease
Let A be the event that a person is tested positive for that disease.
P(D|A) = Probability that someone has a disease given that he tests positive.
We are given that There is an excellent test for the disease; 98.8% of the people with the disease test positive
So, P(A|D)=probability that a person is tested positive given he has a disease = 0.988
We are also given that one person in 10,000 people has a rare genetic disease.
So,![P(D)=\frac{1}{10000}](https://tex.z-dn.net/?f=P%28D%29%3D%5Cfrac%7B1%7D%7B10000%7D)
Only 0.4% of the people who don't have it test positive.
= probability that a person is tested positive given he don't have a disease = 0.004
![P(D^c)=1-\frac{1}{10000}](https://tex.z-dn.net/?f=P%28D%5Ec%29%3D1-%5Cfrac%7B1%7D%7B10000%7D)
Formula:![P(D|A)=\frac{P(A|D)P(D)}{P(A|D)P(D^c)+P(A|D^c)P(D^c)}](https://tex.z-dn.net/?f=P%28D%7CA%29%3D%5Cfrac%7BP%28A%7CD%29P%28D%29%7D%7BP%28A%7CD%29P%28D%5Ec%29%2BP%28A%7CD%5Ec%29P%28D%5Ec%29%7D)
![P(D|A)=\frac{0.988 \times \frac{1}{10000}}{0.988 \times (1-\frac{1}{10000}))+0.004 \times (1-\frac{1}{10000})}](https://tex.z-dn.net/?f=P%28D%7CA%29%3D%5Cfrac%7B0.988%20%5Ctimes%20%5Cfrac%7B1%7D%7B10000%7D%7D%7B0.988%20%5Ctimes%20%281-%5Cfrac%7B1%7D%7B10000%7D%29%29%2B0.004%20%5Ctimes%20%281-%5Cfrac%7B1%7D%7B10000%7D%29%7D)
P(D|A)=
=0.9995
P(D|A)=![0.9995](https://tex.z-dn.net/?f=0.9995)
A)The probability that someone who tests positive has the disease is 0.9995
(B)
=probability that someone does not have disease given that he tests negative
=probability that a person tests negative given that he does not have disease =1-0.004
=0.996
=probability that a person tests negative given that he has a disease =1-0.988=0.012
Formula: ![P(D^c|A^c)=\frac{P(A^c|D^c)P(D^c)}{P(A^c|D^c)P(D^c)+P(A^c|D)P(D)}](https://tex.z-dn.net/?f=P%28D%5Ec%7CA%5Ec%29%3D%5Cfrac%7BP%28A%5Ec%7CD%5Ec%29P%28D%5Ec%29%7D%7BP%28A%5Ec%7CD%5Ec%29P%28D%5Ec%29%2BP%28A%5Ec%7CD%29P%28D%29%7D)
![P(D^c|A^c)=\frac{0.996 \times (1-\frac{1}{10000})}{0.996 \times (1-\frac{1}{10000})+0.012 \times \frac{1}{1000}}](https://tex.z-dn.net/?f=P%28D%5Ec%7CA%5Ec%29%3D%5Cfrac%7B0.996%20%5Ctimes%20%281-%5Cfrac%7B1%7D%7B10000%7D%29%7D%7B0.996%20%5Ctimes%20%281-%5Cfrac%7B1%7D%7B10000%7D%29%2B0.012%20%5Ctimes%20%5Cfrac%7B1%7D%7B1000%7D%7D)
![P(D^c|A^c)=0.99999](https://tex.z-dn.net/?f=P%28D%5Ec%7CA%5Ec%29%3D0.99999)
B)The probability that someone who tests negative does not have the disease is 0.99999