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6. (20 pts) 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).
Testflu Yes No Total
positive
negative
Total
(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?
Professor Halleck i do not understand this problem at all. Could you please explain how to solve this one?
A. 4% known ->98% positive, 2% negative 96% unknown -> 3% positive, 97% negative
(.04*.98)+(.96*.03)=.068 positive on the test for lyme’s disease?
And i forgot how to do part B
.04 = patients who have Lyme’s disease
.96= patients who does’t have Lyme’s disease
Test Lyme’s disease Yes No Total
Positive .98 .02 1
Negative .03 .97 1
Total 1.01 .99 2
A) the probality that the test comes out positive is : ( .4*.98)+(.96*.03) = .068
B) the probability a person has Lyme’s disease given positive test is :
(.4*.98)/((.4*.98)+(.96*.03))= .576
according to the quesiton, 98% of those with disease have positive, which means the rest of 2% is negative..so i think u you did something wrong in your table!!
You made a mistake when placing .4, this will give you be 40 percent. You would want 4 percent so it will be .04 giving you 4 percent =)
You are right, I see my mistake .
Okay, I used a tree to solve this problem. Perhaps it will be easier to understand it that way since we often use it in class. Cheerioooossss
LOL pardon the “Lime” and the -_- face. 15 hours to Exam time O.o
still not clear. the question itself is a bit confusing. Feels like it is missing some thing or written with some mistake. Professor were are you when we need you?
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