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Math Help - [SOLVED] Probability problem..please help!

  1. #1
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    [SOLVED] Probability problem..please help!

    Hey everyone..this is my firts post..I can't solve this problem..can somebody help me pls..

    1.) Assume that a test to detect a disease whose prevalence is (1/1000) has a false positive rate of 5% and a true positive rate of 100%. What is the probability that a person found to have a positive result actually has the disease, assuming that you know nothing about the person's symptoms?

    Define events: D= the event that the person has the disease
    T= the event that the test result is positive.
    Given:

    P(D)=0.001
    P(T|D complement)=0.05
    P(T|D)=1.00


    please can anybody help..i'll really appreciate it..
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by takeyourmark View Post
    Hey everyone..this is my firts post..I can't solve this problem..can somebody help me pls..

    1.) Assume that a test to detect a disease whose prevalence is (1/1000) has a false positive rate of 5% and a true positive rate of 100%. What is the probability that a person found to have a positive result actually has the disease, assuming that you know nothing about the person's symptoms?

    Define events: D= the event that the person has the disease
    T= the event that the test result is positive.
    Given:

    P(D)=0.001
    P(T|D complement)=0.05
    P(T|D)=1.00


    please can anybody help..i'll really appreciate it..
    So you're looking for P(D|T)

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

    But we know that \mathbb{P}(D \cap T)=\mathbb{P}(T|D)\mathbb{P}(D)

    So \mathbb{P}(D|T)=\frac{\mathbb{P}(T|D)\mathbb{P}(D)  }{\mathbb{P}(T)}

    Now what is \mathbb{P}(T) ?
    T=(T \cap D) \cup (T \cap D^c) (because D and its complementary D^c are disjoint)
    And we also have (T \cap D) \cap (T \cap D^c)= \emptyset

    Hence \mathbb{P}(T)=\mathbb{P}(T \cap D)+\mathbb{P}(T \cap D^c)
    =\mathbb{P}(T|D)P(D)+\mathbb{P}(T|D^c)P(D^c)


    Therefore, \boxed{\mathbb{P}(D|T)=\frac{\mathbb{P}(T|D) \mathbb{P}(D)}{\mathbb{P}(T|D)P(D)+\mathbb{P}(T|D^  c)P(D^c)}}<br />
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    So you're looking for P(D|T)

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

    But we know that \mathbb{P}(D \cap T)=\mathbb{P}(T|D)\mathbb{P}(D)

    So \mathbb{P}(D|T)=\frac{\mathbb{P}(T|D)\mathbb{P}(D)  }{\mathbb{P}(T)}

    Now what is \mathbb{P}(T) ?
    T=(T \cap D) \cup (T \cap D^c) (because D and its complementary D^c are disjoint)
    And we also have (T \cap D) \cap (T \cap D^c)= \emptyset

    Hence \mathbb{P}(T)=\mathbb{P}(T \cap D)+\mathbb{P}(T \cap D^c)
    =\mathbb{P}(T|D)P(D)+\mathbb{P}(T|D^c)P(D^c)


    Therefore, \boxed{\mathbb{P}(D|T)=\frac{\mathbb{P}(T|D) \mathbb{P}(D)}{\mathbb{P}(T|D)P(D)+\mathbb{P}(T|D^  c)P(D^c)}}<br />
    Thank you! Thank you so much! Probabilities really confuses me..I don't know what I'll do without your help..but what is the value of P(D^c)? is it equals to 1-P(D) or not? sorry i'm just weak when it comes to probabilities..thank you again!

    i went through my notes..i got it..thanks again!
    Last edited by mr fantastic; February 14th 2009 at 04:25 AM. Reason: Merged posts
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