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Math Help - Test for a certain disease

  1. #1
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    Test for a certain disease

    Testing for a disease. has 0.02 probability of giving a false-positive reading and probability of 0.03 of giving a false-negative result. (a false-positive means testing postive for the disease when it's a healthy person).
    5 people are test, two of whom have the disease and three who don't. X=number of postive readings.

    I am going to assume this is a binomial distribution and there are only 2 outcomes (postive or negative for disease). Right?

    What im having trouble with is the probability that at most 2 of the 5 tests results are positive.. I am confused where to start this
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  2. #2
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    Quote Originally Posted by mightydog78 View Post
    Testing for a disease. has 0.02 probability of giving a false-positive reading and probability of 0.03 of giving a false-negative result. (a false-positive means testing postive for the disease when it's a healthy person).
    5 people are test, two of whom have the disease and three who don't. X=number of postive readings.

    I am going to assume this is a binomial distribution and there are only 2 outcomes (postive or negative for disease). Right?

    What im having trouble with is the probability that at most 2 of the 5 tests results are positive.. I am confused where to start this
    X ~ Binomial(n = 5, p = ?)

    So your first job is to calculate the value of p, the probability of a positive result in a single trial.

    Then calculate \Pr(X \leq 2).
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  3. #3
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    P(randomly chosen person gives + test) = 0.4*0.97 + 0.6*0.02 = 0.4 so is this my p value??

    P(X≤2) = P(X=0) + P(X=1) + P(X=2) = (0.6)⁵+ 5C1*(0.4)*(0.6)⁴+ 5C2*(0.4)*(0.6)= 0.6826

    and then do i go to the binomial tables and find where 0.6826 and 0.4 intersect??
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  4. #4
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    wow im dumb, i calculated the probability haha
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