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 If your blood test is positive, what is the chance that you are a carrier of the disease?

Explanation / Answer

We have been given all the data we need, this is how I will solve the problem:

First solve for - P[person will have a positive blood test] = ?

Definitions have been given of C and P events as: C is event that a randomly selected person is a carrier of the disease. and P is the event that the blood test is positive.

Chance that you are a carrier of the disease, if your blood test is +ive = P(C)*P(P|C)+ P(C')*P(P|C')

= (0.01*0.97) + (0.99*0.06)

= 0.0097 + 0.0594

= 0.0691

If your blood test is positive, what is the chance that you are a carrier of the disease = ?

= P[C | P] = 0.01*0.97 / 0.0691

= 0.1404

Answer is .1404

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