Statistics with Probability

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  • #12965

    Ezra Halleck
    Participant

    5. 4% of a clinic’s patients are known to have Lyme’s disease. A test is developed that is positive in 98% of patients with Lyme’s disease, but it is also positive in 3% of patients who do not have disease. Fill in the following table (use all available digits).

    (a) What is probability that test comes out positive for Lyme’s disease?
    (b) What is probability that person has Lyme’s disease given positive test?

    #16186

    vlad
    Participant

    b) A the event the person has lyme’s disease and B the event thes is +

    P(A|B) = (.98)(.04)/(.98)(.04)+(.03)(.96)
    = .576
    there’s 58% chance that a patient who tested + actually has lyme’s disease.

    right? wrong?

    #16196

    Ezra Halleck
    Participant

    I think you got it. Just be careful about how you write down the calculation. There should be an extra set of parentheses:
    P(A|B) = (.98)(.04)/((.98)(.04)+(.03)(.96))

    #16230

    Afzal
    Participant

    a)
    P(L)=0.04
    P(L^c)=0.96
    P (+|L)=0.98
    P(L^c)=0.03

    P(+)=P (+|L)*P(L)+(P(+|L^c)*P(L^c))
    =(0.98)*(0.04)+(0.03)*(0.96)
    =0.068
    =6.8%
    so, 6.8% probability that test comes out positive for Lyme’s disease.

    #16241

    well for this question solutions i had was

    a) no because 3/3 joints

    b) p(a)=1/6 p(b)=1/6 and p(a and b)=1/36

    c)p(a) p(b)=p(a N b) yes true

    #16281

    im confused about what part b of the question is asking. i got 6.8 percent for a, is b 3.9%? if not how do you get b.

    #16282

    ok i got b 57 % thanks

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