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Math Help - False positives

  1. #1
    Junior Member nimon's Avatar
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    False positives

    I'm quite happy to answer part 1. of the following problem.

    "You are diagnosed with an uncommon disease. You know that there is only a 1% chance of getting it. Let D stand for the event 'you have the disease' and T stand for 'the test says so'. It is known that the test is imperfect: P(T | D) = 0.98 and P(T^{c} | D^{c}) = 0.95.

    1. Given that you test positive, what is the probability that you really have the disease?
    2. You obtain a second opinion: an independent repetition of the test. You test positive again. Given this, what is the probability that you really have the disease?"

    So, for part 1. I use Bayes' theorem to get
     P(D|T) = \frac{P(T|D)P(D)}{P(T|D)P(D) + P(T|D^{c})P(D^{c})} = \frac{0.98 \times 0.01}{0.98 \times 0.01 + 0.05 \times 0.99} = 0.165
    which is correct, according to my book. However, when I apply similar reasoning to the second scenario I get an impossible answer.
     P(D|T_{1} \cap T_{2}) = \frac{ P(T_{1} \cap T_{2} | D)P(D) }{ P(T_{1}\cap T_{2}) } = \ldots
    This is where I get stuck. I know that the independence of T_{1} and T_{2} doesn't necessarily imply their conditional independence (conditioned on D) but I have no idea how to proceed. If I just assume the conditional independence I get an answer greater than 1, which is obviously wrong.
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  2. #2
    MHF Contributor
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    Re: False positives

    Hey nimon.

    If T1 and T2 are independent then

    P(D|T1 and T2)
    = P(T1 and T2 and D)/P(T1 and T2)
    = P(T1 and D)*P(T2 and D)/[P(T1)*P(T2)]
    = P(D|T1)*P(D|T2)

    This should make sense intuitively since both processes are independent and don't change given data from the other process.
    Thanks from nimon
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  3. #3
    Junior Member nimon's Avatar
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    Re: False positives

    Thanks for your input Chiro.

    If I understand, then I should have P(D|T_{1} \cap T_{2}) = P(D|T_{1})P(D|T_{2}) < P(D|T_{1}).

    Now this makes no sense to me: if I get a second positive, shouldn't the probability of my having the disease go up, and not down?

    Furthermore, why do you have P(D\cap T_{1} \cap T_{2}) = P(D \cap T_{1})P(D \cap T_{2})? This is something I tried and then investigated; I think I can come up with lots of examples where T_{1} and T_{2} are independent but where T_{1} \cap D and T_{2} \cap D are not.

    This much is contained in the opening post, where I say that the independence of two events does not imply their independence when conditioned on some third event, and vice-versa.

    I seem to be going in circles...
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  4. #4
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    Re: False positives

    You wanted the intersection of events and not the union (in your post).

    If you want the union then use the fact P(A OR B) = P(A) + P(B) - P(A and B).
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  5. #5
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    Re: False positives

    This problem is already solved: here
    Thanks from nimon
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  6. #6
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    Re: False positives

    Here's how I would do this problem. Imagine a population of 10000 people. "There is only a 1% chance of getting it." so 1% of 10000= 100 people have the disease and 10000-100= 9900 do not. 98% of the people who have the disease test positive so of the 1000 who have the disease 98 test positive and 2 do not. Since 95% of the people who do not have the disease do not test positive, 5% of 9900= 495 people who do NOT have the disease test positive.

    1.Given that you test positive, what is the probability that you really have the disease?
    Of the 495+98= 593 people who test positive, 98 of them have the disease so the probability that you have the disease, given that the test was positive is 98/593= 16%. The probablity that you do NOT have the disease, given that the test was positive, is 100- 16= 84%.

    2.You obtain a second opinion: an independent repetition of the test. You test positive again. Given this, what is the probability that you really have the disease?
    The probability is now 1.00- .84^2= 30%
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