Respuesta :
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 [tex]D^c[/tex] 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,[tex]P(D)=\frac{1}{10000}[/tex]
Only 0.4% of the people who don't have it test positive.
[tex]P(A|D^c)[/tex] = probability that a person is tested positive given he don't have a disease = 0.004
[tex]P(D^c)=1-\frac{1}{10000}[/tex]
Formula:[tex]P(D|A)=\frac{P(A|D)P(D)}{P(A|D)P(D^c)+P(A|D^c)P(D^c)}[/tex]
[tex]P(D|A)=\frac{0.988 \times \frac{1}{10000}}{0.988 \times (1-\frac{1}{10000}))+0.004 \times (1-\frac{1}{10000})}[/tex]
P(D|A)=[tex]\frac{2470}{2471}[/tex]=0.9995
P(D|A)=[tex]0.9995[/tex]
A)The probability that someone who tests positive has the disease is 0.9995
(B)
[tex]P(D^c|A^c)[/tex]=probability that someone does not have disease given that he tests negative
[tex]P(A^c|D^c)[/tex]=probability that a person tests negative given that he does not have disease =1-0.004
=0.996
[tex]P(A^c|D)[/tex]=probability that a person tests negative given that he has a disease =1-0.988=0.012
Formula: [tex]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)}[/tex]
[tex]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}}[/tex]
[tex]P(D^c|A^c)=0.99999[/tex]
B)The probability that someone who tests negative does not have the disease is 0.99999
