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Math Help - Bayes Rule probability problem.

  1. #1
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    Bayes Rule probability problem.

    Hi all,

    I am not sure if this belongs to University Math help. It looks like it's basic enough but we are doing it in college. I hope I am not intruding in the wrong territory.

    The problem is as follows:

    A certain disease can be detected by a blood test in 95% of those who have it. Unfortunately, the test also has a 0.02 probability of showing that a person has the disease when in fact he or she does not. It has been estimated that 1% of those people who are routinely tested actually have the disease. If the test shows that a certain person has the disease, find the probability that the person actually has it.


    This is how I did it:

    For straight forwardness we will look at the population as being 10000 people.

    95% of those who have it will test "yes" i.e. 95% of 1% = 0.95 x 0.01 = 0.0095 from all population, so from 10000 people it would be 95 people

    2% of those who do not have it will also test "yes" i.e. 2% of 99% = 0.02 x 0.99 = 0.0198 from all population, so from 10000 people it would be 198 people

    So, the total number of all "yes" tests (positive "yes" and negative "yes") would be 95+198=293 "yes" tests

    Since we know that 1% of population actually have the disease i.e. 100 people then we can conclude that chance of having disease if the test was positive is 100/293=0.3413

    As you can see, my answer came out as 0.3413. Is that correct? The problem is - my lecturer's answer is 0.324


    Thank you in advance!
    Last edited by mr fantastic; May 9th 2011 at 12:30 PM. Reason: Title.
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  2. #2
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    Quote Originally Posted by johammbass View Post
    Hi all,

    I am not sure if this belongs to University Math help. It looks like it's basic enough but we are doing it in college. I hope I am not intruding in the wrong territory.

    The problem is as follows:

    A certain disease can be detected by a blood test in 95% of those who have it. Unfortunately, the test also has a 0.02 probability of showing that a person has the disease when in fact he or she does not. It has been estimated that 1% of those people who are routinely tested actually have the disease. If the test shows that a certain person has the disease, find the probability that the person actually has it.


    This is how I did it:

    For straight forwardness we will look at the population as being 10000 people.

    95% of those who have it will test "yes" i.e. 95% of 1% = 0.95 x 0.01 = 0.0095 from all population, so from 10000 people it would be 95 people

    2% of those who do not have it will also test "yes" i.e. 2% of 99% = 0.02 x 0.99 = 0.0198 from all population, so from 10000 people it would be 198 people

    So, the total number of all "yes" tests (positive "yes" and negative "yes") would be 95+198=293 "yes" tests

    Since we know that 1% of population actually have the disease i.e. 100 people then we can conclude that chance of having disease if the test was positive is 100/293=0.3413

    As you can see, my answer came out as 0.3413. Is that correct? The problem is - my lecturer's answer is 0.324
    Well I get P(D|+)=\frac{P(+|D)P(D)}{P(+|D)P(D)+P(+|D^c)P(D^c)  }=0.324232082.
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  3. #3
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    Dear Plato,

    If you are not very busy, would you be able to the solution in explicitly because I am not clearly understanding all the terms. Would you be able to tell me where I went wrong with my reasoning?

    Thank you
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  4. #4
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    Quote Originally Posted by johammbass View Post
    would you be able to the solution in explicitly
    You should understand that this is not a tutorial service.
    I will however give you the numbers.
    + means a positive test.
    D means that an individual actually has the disease.
    So P(+|D)=0.95,~P(+|D^c)=0.02,~\&~P(D)=0.01~.
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  5. #5
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    Hmm, I don't know, I thought the answer would have to be the probability of being actually diseased over the probability of a positive test but I guess I was wrong
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  6. #6
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    Quote Originally Posted by johammbass View Post
    Hmm, I don't know, I thought the answer would have to be the probability of being actually diseased over the probability of a positive test but I guess I was wrong
    The question asks, "Given that the test is positive, what is the probability that the disease is actually present"?
    That is P(D|+)=\frac{P(D~\&~+)}{P(+)}.
    One needs to understand conditional probabilies.
    If you have doubts about that, then that may be you problem.
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