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Math Help - Conditional Probability

  1. #1
    Super Member Aryth's Avatar
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    Conditional Probability

    Suppose there is a medical diagnostic test for a disease. The sensitivity of the test is .95. This means that if a person has the disease, the probability that the test gives a positive response is .95. The specificity of the test is .90. This means that if a person does not have the disease, the probability that the test gives a negative response is .90, or that the false positive rate of the test is .10. In the population, 1% of the people have the disease. What is the probability that a person tested has the disease, given the results of the test is positive? Let D e the event "the person has the disease" and let T be the event "the test gives a positive result."

    All I really need to know is how to find P(D). The rest I can handle on my own.

    Thanks.

    The way I solved it was this:

    I need to find P(D|T) = \frac{P(T|D)P(D)}{P(T|D)P(D) + P(T|D')P(D')}

    P(D|T) = \frac{(0.95)(0.01)}{(0.95)(0.01) + (0.10)(0.99)}

    P(D|T) = \frac{0.0095}{0.0095 + 0.099}

    P(D|T) = \frac{0.0095}{0.1085}

    P(D|T) = 0.0875

    Is that right?
    Last edited by Aryth; December 20th 2008 at 06:53 PM.
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  2. #2
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    I attempted the question and used bayes theorem and got the same answer as what you got.
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    Quote Originally Posted by Aryth View Post
    Suppose there is a medical diagnostic test for a disease. The sensitivity of the test is .95. This means that if a person has the disease, the probability that the test gives a positive response is .95. The specificity of the test is .90. This means that if a person does not have the disease, the probability that the test gives a negative response is .90, or that the false positive rate of the test is .10. In the population, 1% of the people have the disease. What is the probability that a person tested has the disease, given the results of the test is positive? Let D e the event "the person has the disease" and let T be the event "the test gives a positive result."

    All I really need to know is how to find P(D). The rest I can handle on my own.

    Thanks.

    The way I solved it was this:

    I need to find P(D|T) = \frac{P(T|D)P(D)}{P(T|D)P(D) + P(T|D')P(D')}

    P(D|T) = \frac{(0.95)(0.01)}{(0.95)(0.01) + (0.10)(0.99)}

    P(D|T) = \frac{0.0095}{0.0095 + 0.099}

    P(D|T) = \frac{0.0095}{0.1085}

    P(D|T) = 0.0875

    Is that right?
    Yes. Well done.
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    Super Member Aryth's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Yes. Well done.

    The reason why I posted this was because I multiplied 0.10 and 0.99 wrong and couldn't arrive at the correct answer. But I do have one more question, the answer in my book provides this:

    P(D|T) = \frac{P(D \cap T)}{P(T)} = 0.0875

    But, what is P(T) without knowing whether or not the person has the disease? Or are you supposed to assume that they have the disease? I tried both scenarios and couldn't arrive at that solution.
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    Quote Originally Posted by Aryth View Post
    The reason why I posted this was because I multiplied 0.10 and 0.99 wrong and couldn't arrive at the correct answer. But I do have one more question, the answer in my book provides this:

    P(D|T) = \frac{P(D \cap T)}{P(T)} = 0.0875

    But, what is P(T) without knowing whether or not the person has the disease? Or are you supposed to assume that they have the disease? I tried both scenarios and couldn't arrive at that solution.

    P(D|T) = \frac{P(T|D)P(D)}{P(T|D)P(D) + P(T|D')P(D')} and P(D|T) = \frac{P(D \cap T)}{P(T)} are identical statements.

    Pr(T) is just the total probability of testing positive and is given by P(T|D)P(D) + P(T|D')P(D'). Think about it ...
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    Super Member Aryth's Avatar
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    I understand that the two are identical, and I understand the law of total probability.

    But the way that the answer is formed in the book makes the solution look immediately obvious from the conditional probability statement alone. Is that the case?
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    Quote Originally Posted by Aryth View Post
    I understand that the two are identical, and I understand the law of total probability.

    But the way that the answer is formed in the book makes the solution look immediately obvious from the conditional probability statement alone. Is that the case?
    Maybe for some ..... But not for me.
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  8. #8
    Super Member Aryth's Avatar
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    Thanks for the help. Its much appreciated.
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