Problem 3. Suppose that 4% of the patients tested in a clinic are infected with

Question

Problem 3. Suppose that 4% of the patients tested in a clinic are infected with avian flue. Also, suppose that when a blood test for avian flue is given, 98% of the patients infected with avian flue test positive and that 2% of the non-infected patients test positive. What is the probability that (a) a patient testing positive for avian flue is infected with it? (b) a patient testing positive for avian flue is not infected with it? (c) a patient testing negative for avian flue is infected with it? (d) a patient testing negative for avian flue is not infected with it? Provide detailed justification for your answers.

Explanation / Answer

let probability of infection is P(I) ,not infected is P(Ic) and probability of tested positive -P(T) and tested negative =P(Tc)

P(T) =P(I)*P(T|I)+P(Ic)*P(T|Ic)=0.04*0.98+(1-0.04)*0.02=0.0584

a)P(infected|tested positive) = P(I)*P(T|I)/P(T) =0.04*0.98/0.0584=0.6712

b)P(not infected|tested positive)=P(Ic)*P(T|Ic)/P(T) =(1-0.04)*0.02/0.0584 =0.3288

c)P(infected|tested negative) =P(I)*P(Tc|I)/P(Tc) =0.04*(1-0.98)/(1-0.0584)=0.00085

d)P(not infected|tested negative) =P(Ic)*P(Tc|Ic)/P(Tc) =0.96*(1-0.02)/(1-0.0584)=0.99915

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