Results 1 to 2 of 2

Math Help - Probability

  1. #1
    Newbie
    Joined
    Dec 2006
    Posts
    23

    Probability

    Hi
    I've got a problem with a question
    Probability-prob-45.jpg
    Thanks for all your help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,707
    Thanks
    627
    Hello, mauro21!

    Suppose there was a cancer diagnostic test that was 95% accurate
    both on those that do and those that do not have the disease.
    If 4% of the population have cancer, compute the probability that a tested person
    has cancer, given that his/her test result indicates so.

    This is a Conditional Probability problem
    . . and requires a lot of "translating".


    Let C mean the person has cancer.
    And ~C mean the person does not have cancer.
    . . Then: .P(C) = 0.04, .P(~C) = 0.96

    Let pos mean the test was positive for cancer.
    Let neg means the test was negative for cancer.


    The test is 95% accurate.

    This means: if a person has cancer, 95% of the time the test will be posistive.
    . . That is: .P(pos | C) = 0.95

    And if a person does not have cancer, 95% of the time the test will be negative.
    . . That is: .P(neg | ~C) = 0.95


    We want the probability that the person has cancer, given that the test is positive.
    . . That is: .P(C | pos)

    . . . . . . . . . . . . . . . . . . . . . . . . P(C and pos)
    Bayes' Theorem: . P(C | pos) . = . ---------------
    . . . . . . . . . . . . . . . . . . . . . . . . . . P(pos)


    For the numerator: .P(C and pos) .= .(0.04)(0.95) .= .0.038


    For the denominator, there are two ways that the test can be positive.

    . . (1) The person has cancer and the test is accurate: (0.04)(0.95) = 0.38

    . . (2) The person does not have cancer and the test is wrong: (0.96)(0.05) = 0.048

    Hence: .P(pos) .= .0.38 + 0.48 .= .0.086


    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.038
    Bayes' Theorem becomes: .P(C | pos) .= .------- . .44%
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.086

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: January 21st 2011, 11:47 AM
  2. Replies: 0
    Last Post: December 6th 2010, 04:57 PM
  3. Replies: 3
    Last Post: May 29th 2010, 07:29 AM
  4. Replies: 1
    Last Post: February 18th 2010, 01:54 AM
  5. Replies: 3
    Last Post: December 15th 2009, 06:30 AM

Search Tags


/mathhelpforum @mathhelpforum