Respuesta :
Answer:
a) [tex] P(+) = 0.98*0.15 + 0.1*0.85 = 0.232[/tex]
b) [tex] P(+ \cap D)= P(+|D) *P(D)= 0.98*0.15=0.147[/tex]
[tex] P(D|+) =\frac{0.147}{0.232}=0.634[/tex]
c) [tex] P(+ \cap ND)= P(+|ND) *P(ND)= 0.1*0.85=0.085[/tex]
[tex] P(ND|+) = \frac{0.085}{0.232}= 0.366[/tex]
Step-by-step explanation:
For this case we define the following notation:
+ represent the event of getting a positive result
D = represent the event of having the disease
ND = represent the event of NO having the disease
From the info given we know that:
[tex] P(+|D) = 0.98 , P(+|ND) = 0.1 , P(D) = 0.15[/tex]
And by the complement rule we can find:
[tex] P(ND) = 1-P(D) = 1-0.15=0.85[/tex]
Part a
For this case we want to find this probability P(+) and for this case we can use the Bayes total rule and we can do this:
[tex] P(+) = P(+|D) P(D) + P(+|ND) P(ND)[/tex]
And if we replace we got:
[tex] P(+) = 0.98*0.15 + 0.1*0.85 = 0.232[/tex]
Part b
We want to find this probability [tex] P(D|+)[/tex] and using the bayes rule we have:
[tex] P(D|+) = \frac{P(+\ cap D)}{P(+)}[/tex]
We can find the numerator from:
[tex] P(+|D) = \frac{P(+ \cap D)}{P(D)}[/tex]
[tex] P(+ \cap D)= P(+|D) *P(D)= 0.98*0.15=0.147[/tex]
And then if we replace we got:
[tex] P(D|+) =\frac{0.147}{0.232}=0.634[/tex]
Part c
We want to find this probability [tex] P(ND|+)[/tex] and using the bayes rule we have:
[tex] P(ND|+) = \frac{P(+\ cap ND)}{P(+)}[/tex]
We can find the numerator from:
[tex] P(+|ND) = \frac{P(+ \cap ND)}{P(ND)}[/tex]
[tex] P(+ \cap ND)= P(+|ND) *P(ND)= 0.1*0.85=0.085[/tex]
And then if we replace we got:
[tex] P(ND|+) = \frac{0.085}{0.232}= 0.366[/tex]
