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 negative, what is the chance that you are not a carrier of the disease? (Be sure to use at least four significant digits.) Correct

Explanation / Answer

Here as C denote the event that a randomly selected person is a carrier of the disease., therefore we are given that:

P(C) = 0.01

Also as P denote the event that the blood test is positive, we are given that:

P( P | C) = 0.97 and therefore P( N | C) = 1 - 0.97 = 0.03 where N stands for negative test.

Also we are given that:

P( P | C') = 0.06 as there is a 6% chance of giving a positive test result falsely indicating that a healthy person

This means that P( N | C') = 1 - 0.06 = 0.94

Using the last of total probability, the probability that the test is negative is computed as:

P(N) = P( N | C)P(C) + P( N | C')P(C') = 0.03*0.01 + 0.94*(1 - 0.01) = 0.9309

Now given that the test is negative, probability that you are not carrier of the disease is computed as: (Using bayes theorem ) :

P( C' | N) = P( N | C')P(C') / P(N) = 0.94*0.99 / 0.9309 = 0.9997

Therefore 0.9997 is the required probability here.

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