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Math Help - conditional probability help

  1. #1
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    conditional probability help

    Hi, I was wondering if anyone could help me out with where to get started on this prblem


    Bad gums may mean a bad heart. Researchers discovered that 78% of people who have suffered a heart attack had periodontal disease, an inflammation of the gums. Only 30% of healthy people have this disease. Suppose that in a certain community heart attacks are quite rare, occurring with only 11% probability.

    A. If someone has periodontal disease, what is the probability that he or she will have a heart attack?


    B. If 40% of the people in a community will have a heart attack, what is the probability that a person with periodontal disease will have a heart attack?
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  2. #2
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    Quote Originally Posted by jason0 View Post
    Hi, I was wondering if anyone could help me out with where to get started on this prblem


    Bad gums may mean a bad heart. Researchers discovered that 78% of people who have suffered a heart attack had periodontal disease, an inflammation of the gums. Only 30% of healthy people have this disease. Suppose that in a certain community heart attacks are quite rare, occurring with only 11% probability.

    A. If someone has periodontal disease, what is the probability that he or she will have a heart attack?


    B. If 40% of the people in a community will have a heart attack, what is the probability that a person with periodontal disease will have a heart attack?
    Draw a tree diagram to see what's happening. The first two branches are Heart attack (H) and no Heart attack (H'). Pr(H) = 0.11, Pr(H') = 0.89.

    The two branches that go from H and H' are Dental disease (D) and no Dental disease (D'). Pr(D | H) = 0.78 and so Pr(D' | H) = 0.22. Also Pr(D | H') = 0.3 and so Pr(D' | H') = 0.7.

    For part A the tree diagram shuold make it clear how to calculate Pr(H | D).

    For part B, you first need to edit the tree diagram because now Pr(H) = 0.4.
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    Quote Originally Posted by mr fantastic View Post
    Draw a tree diagram to see what's happening. The first two branches are Heart attack (H) and no Heart attack (H'). Pr(H) = 0.11, Pr(H') = 0.89.

    The two branches that go from H and H' are Dental disease (D) and no Dental disease (D'). Pr(D | H) = 0.78 and so Pr(D' | H) = 0.22. Also Pr(D | H') = 0.3 and so Pr(D' | H') = 0.7.

    For part A the tree diagram shuold make it clear how to calculate Pr(H | D).

    For part B, you first need to edit the tree diagram because now Pr(H) = 0.4.
    wouldn't that make the answer for A just .11*.78? because that's not right and i'm still confused
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    Quote Originally Posted by jason0 View Post
    wouldn't that make the answer for A just .11*.78? because that's not right and i'm still confused
    Do you understand conditional probability at all? I suggest you thoroughly review it. \Pr(H | D) = \frac{\Pr(H \cap D)}{\Pr(D)}.

    From the tree diagram \Pr(D) = (0.11)(0.78) + (0.89)(0.3) = .... and \Pr(H \cap D) = (0.11)(0.78) = .....
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