Show My Work state medical school has discovered a new test for tuberclosa. ( th

Question

Show My Work state medical school has discovered a new test for tuberclosa. ( the test naaas a person beats probability of a positive test is O.81 tuberculosis. Assume that in the general populatien, the probablity that a person has tubercuiosis is 0.06 following categories. (Enter your answers to four decimal places) has tuberculosis, the test is pesitive.) Experimentation has shown that the 1, given that a person has tuberculosis. The probability is o.o9 that the test registers postive, given that the person does not have . what i·th. probabley that . penon chosen at rar dem fali in (a) have tuberculosis and have a positive test (b) not have tuberculosis (e) not have tuberculosis and have a positive test Need Help? Show My Work in 888a 2 3

Explanation / Answer

Solution :

We are given that P(positive test| Having tuberculosis) = 0.81

P(Having tuberculosis) = 0.06 and P( positive test| Not having tuberculosis) = 0.09

a) To find Probability that person have tuberculosis and positive test.

Using the conditional probability formula ,P(A|B) = P(A and B)/P(B)

=>P(A and B) = P(A|B)*P(B)

So here P( Having Tuberculosis and Positive test) = P(Positive test| Having tuberculosis) * P( having tuberculosis)

= 0.81*0.06 =0.0486

Therefore, Probability that person have tuberculosis and positive test = 0.0486

b) Probability that person choose at random do not have tuberculosis = 1 - P( having Tuberculosis) = 1- 0.06 = 0.94

c) Probability that person not have tuberculosis and test positive is same as P( test positive and not having tuberculosis)

Again using conditional probability formula, we get

P( test positive| not having tuberculosis) = P( test positive and not having tuberculosis)/P(not having tuberculosis)

So,P( test positive and not having tuberculosis)= P( test positive| not having tuberculosis)*P(not having tuberculosis)

= 0.09*0.94 = 0.0846

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