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Math Help - Baye's Theorem question

  1. #1
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    Baye's Theorem question

    the question goes as follows:

    Medical case histories indicate that different illnesses may produce identical symptoms. Suppose that a particular set of symptoms, denoted H, occurs only when one of the tress illnesses,  I_1,\ 1_2, \ I_3 occurs. Assume that the simultaneous occurrence of more ten one of these illnesses is impossible and that

     P(I_1) = 0.01, \ P(I_2) = 0.005, \ P(I_3) = 0.02

    in addition:

     P(H|I_1) = 0.9, \ P(H|I_2) = 0.95, \ P(H|I_3) = 0.75

    assuming that an ill person exhibits the symptoms, H, what is the probability that the person has illness I_1?

    solution:

    P(I_1|H) = \ ?

     P(I_1|H) = \frac{H \cap I_1}{P(H)} \ = \  \frac{(H|I_1)(I_1)}{P(H)}

    To find  P(H) = \left( P(H \cap I_1) \ \bigcup \ P(H \cap \overline{I_1}) \right) \ \bigcup \ \left( P(H \cap I_2) \bigcup \ P(H \cap \overline{I_2}) \right) \ \bigcup \left(P(H \cap I_3) \ \bigcup \ P(H \cap \overline{I_3}) \right)

    so P(H) = (0.9)(0.01) +0 +(0.95)(0.05) +0+(0.75)(0.02) +0 = 0.1525

    therefore:  P(I_1|H) = \frac{(0.9)(0.1)}{0.1525} = 0.590164

    is this correct, or am I missing something?
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  2. #2
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    Quote Originally Posted by lllll View Post
    Medical case histories indicate that different illnesses may produce identical symptoms. Suppose that a particular set of symptoms, denoted H, occurs only when one of the tress illnesses,  I_1,\ 1_2, \ I_3 occurs. Assume that the simultaneous occurrence of more than one of these illnesses is impossible and that
     P(I_1) = 0.01, \ P(I_2) = 0.005, \ P(I_3) = 0.02
    in addition:
     P(H|I_1) = 0.9, \ P(H|I_2) = 0.95, \ P(H|I_3) = 0.75
    “Assume that the simultaneous occurrence of more than one of these illnesses is impossible” means that P(H) = P\left( {H \cap I_1 } \right) + P\left( {H \cap I_2 } \right) + P\left( {H \cap I_3 } \right).

    From the given you can find each of these: P\left( {H \cap I_1 } \right) = P\left( {H|I_1 } \right)P\left( {I_1 } \right)
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