Let C denote the event that a randomly selected person is a carrier of the disea

Question

Let C denote the event that a randomly selected person is a carrier of the disease. Let P denote the event that the blood test is positive. Suppose it is known that 1% of the population are carriers of a particular disease. A blood test has a 97% chance of giving a positive test result correctly identifying carriers of the disease, but also has a 6% chance of giving a positive test result falsely indicating that a healthy person (i.e. non-carrier) has the disease. Thus, P(C) = .01, P(P|C) = .97, and P(P|C’) = .06 What is the probability that a person will have a positive blood test?

Explanation / Answer

So C denotes the event that a randomly selected person is a carrier of the disease.

And, P denotes the event that the blood test is positive.

P[person will have a positive blood test] = P(C)*P(P|C)+ P(P|C')*P(C')

= 0.01*0.97 + 0.99*0.06

= 0.0097 + 0.0594

= 0.0691

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