[Ezra Halleck]

5. 4% of a clinic’s patients are known to have Lyme’s disease. A test is developed that is positive in 98% of patients with Lyme’s disease, but it is also positive in 3% of patients who do not have disease. Fill in the following table (use all available digits).

(a) What is probability that test comes out positive for Lyme’s disease?
(b) What is probability that person has Lyme’s disease given positive test?

[vlad]

b) A the event the person has lyme’s disease and B the event thes is +

P(A|B) = (.98)(.04)/(.98)(.04)+(.03)(.96)
= .576
there’s 58% chance that a patient who tested + actually has lyme’s disease.

right? wrong?

[Ezra Halleck]

I think you got it. Just be careful about how you write down the calculation. There should be an extra set of parentheses:
P(A|B) = (.98)(.04)/((.98)(.04)+(.03)(.96))

[Afzal]

a)
P(L)=0.04
P(L^c)=0.96
P (+|L)=0.98
P(L^c)=0.03

P(+)=P (+|L)*P(L)+(P(+|L^c)*P(L^c))
=(0.98)*(0.04)+(0.03)*(0.96)
=0.068
=6.8%
so, 6.8% probability that test comes out positive for Lyme’s disease.

[Nikoy O’Gilvie]

well for this question solutions i had was

a) no because 3/3 joints

b) p(a)=1/6 p(b)=1/6 and p(a and b)=1/36

c)p(a) p(b)=p(a N b) yes true

[Richard Matos]

im confused about what part b of the question is asking. i got 6.8 percent for a, is b 3.9%? if not how do you get b.

[Richard Matos]

ok i got b 57 % thanks