Results 1 to 2 of 2

Thread: Baye's Theorem question

  1. #1
    Senior Member
    Joined
    Jan 2008
    From
    Montreal
    Posts
    311
    Awards
    1

    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 $\displaystyle H$, occurs only when one of the tress illnesses, $\displaystyle I_1,\ 1_2, \ I_3$ occurs. Assume that the simultaneous occurrence of more ten one of these illnesses is impossible and that

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

    in addition:

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

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

    solution:

    $\displaystyle P(I_1|H) = \ ?$

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

    To find $\displaystyle 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 $ $\displaystyle \left(P(H \cap I_3) \ \bigcup \ P(H \cap \overline{I_3}) \right)$

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

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

    is this correct, or am I missing something?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,741
    Thanks
    2812
    Awards
    1
    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 $\displaystyle H$, occurs only when one of the tress illnesses, $\displaystyle I_1,\ 1_2, \ I_3$ occurs. Assume that the simultaneous occurrence of more than one of these illnesses is impossible and that
    $\displaystyle P(I_1) = 0.01, \ P(I_2) = 0.005, \ P(I_3) = 0.02$
    in addition:
    $\displaystyle P(H|I_1) = 0.9, \ P(H|I_2) = 0.95, \ P(H|I_3) = 0.75$
    “Assume that the simultaneous occurrence of more than one of these illnesses is impossible” means that $\displaystyle P(H) = P\left( {H \cap I_1 } \right) + P\left( {H \cap I_2 } \right) + P\left( {H \cap I_3 } \right).$

    From the given you can find each of these: $\displaystyle P\left( {H \cap I_1 } \right) = P\left( {H|I_1 } \right)P\left( {I_1 } \right)$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Statistical probility, baye's theorem
    Posted in the Statistics Forum
    Replies: 7
    Last Post: Dec 28th 2011, 06:44 AM
  2. Baye's theorem
    Posted in the Statistics Forum
    Replies: 4
    Last Post: Dec 27th 2011, 08:13 AM
  3. Conditional Probability / Baye's Thm
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: Jul 23rd 2011, 10:43 PM
  4. [SOLVED] Baye's Rule 2
    Posted in the Statistics Forum
    Replies: 1
    Last Post: Feb 10th 2011, 01:37 PM
  5. [SOLVED] Baye's Rule
    Posted in the Statistics Forum
    Replies: 0
    Last Post: Feb 10th 2011, 12:20 PM

Search Tags


/mathhelpforum @mathhelpforum