Quiz 6: First A Level Question
AZobi
A certain disease has a prevalence of one percent in a population. A test for the disease has a false positive rate of .05 and a false negative rate of .08. The false positive and false negative rates depend only on whether or not a person has the disease. A person tests positive for the disease and then takes the test one more time. What is the probability that the person will test positive on the second test?
AZobi: March 4, 2015, 2:29 a.m.
Once again, I am not 100% sure and would really appreciate everyone's input. This is my answer: [(.85)(.99)(.01)]/[(.85)(.99)(.01)+(.02)(.2)(.99)]. I will be more than glad to explain how I got these answers.
dynguyen: March 5, 2015, 1:35 a.m.
Hi! I got that answer for the B level question. I think you switched your answers.
AZobi: March 5, 2015, 2:17 a.m.
Here is my A answer: [(.05)(.05)(.99)+(.92)(.92)(.01)]/[(.92)(.92)(.01)+(.08)(.92)(.01)+(.05)(.05)(.99)+(.95)(.05)(.99)]. Sorry about the mix-up!
ninafukuma: March 5, 2015, 6:21 p.m.
Could you explain how you got the denominator? I did it as P(+ on 2nd test given + on 1st test), so (P + on 2nd and P + on 1st)/(P + on 1st)
martha2A: March 5, 2015, 8:20 p.m.
I got the same answer as you for the numerator but not for the denominator. I got [(.01)(.92)(.92)+(.99)(.05)(.05)]/[(.01)(.92)+(.99)(.05)]. My numerator is the probability that a person tests positive on the first test and my numerator is the probability that a person test positive on both tests.
martha2A: March 5, 2015, 8:23 p.m.
The first numerator on the last sentence should be denominator. Sorry about that.
AZobi: March 5, 2015, 8:41 p.m.
I think I am wrong, but I included the total outcomes that could result after testing positive. In other words, I thought getting a negative on the second test was part of the denominator too. I can see why your answer makes more sense
martha2A: March 6, 2015, 9:20 a.m.
Oh I see. That approach would work for regular probability questions, but for this question we use Bayes' Theorem because we are given the information that the person already tested positive on the first test.
AZobi: March 6, 2015, 11:56 a.m.
Thank you Martha! :)
NatalieNguyen2F: March 7, 2015, 9:04 p.m.
Both denominators actually give the same numerical value so it doesn't matter how you think of it!
Alan_Mendoza: March 7, 2015, 10:34 p.m.
Personally, it was simpler to think of the denominator as 1-P(testing negative on first test). Could help out if you don't understand the other way.
weisbart: March 8, 2015, 1:02 p.m.
Martha, you rock! Yes, Alan, you could definitely compute the denominator this way. Sometimes this will make a big difference in difficulty. In this case, I think it's a matter of preference, but that's just an opinion. \[\] Try solving this problem in a slightly different way too to see a different perspective. You could have the disease or not. If you have the disease, there is a .92 chance of testing positive. If you don't, you have a .05 chance of testing positive. Now, what is the probability that you have and don't have the disease? You have the information that you tested positive on the first test, so this should let you know the probability of having or not having the disease (use Bayes' Theorem). You will get the same answer as Martha does, but it will look just slightly different. Anyone want to try it?