# Statistics with Probability

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• FinalReview 5
• #12965

Ezra Halleck
Participant

5. 4% of a clinicâ€™s patients are known to have Lymeâ€™s disease. A test is developed that is positive in 98% of patients with Lymeâ€™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 Lymeâ€™s disease?
(b) What is probability that person has Lymeâ€™s disease given positive test?

#16186

Participant

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?

#16196

Ezra Halleck
Participant

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))

#16230

Afzal
Participant

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 Lymeâ€™s disease.

#16241

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

#16281

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.

#16282

ok i got b 57 % thanks

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