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The state medical school has discovered a new test for tuberculosis. (If the test indicates a person has tuberculosis, the
test is positive.) Experimentation has shown that the probability of a positive test is 0.79, given that a person has
tuberculosis. The probability is 0.08 that the test registers positive, given that the person does not have tuberculosis.
Assume that in the general population, the probability that a person has tuberkulosis is 0.05. What is the probability that a
person chosen at random will fall in the following categories. (Enter your answers to four decimal places.)
(a) have tuberculosis and have a positive test
(b) not have tuberculosis
(c) not have tuberculosis and have a positive test

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Cricetus

The probability according to the given scenario is:

(a) 0.0395

(b) 0.95

(c) 0.076

The given values are:

  • [tex]P(B/A) = 0.79[/tex]
  • [tex]P(A) = 0.05[/tex]

(a)

The probability that the  b/w has tuberculosis as well as having (+) test will be:

→ [tex]P(A \cap B) = P(B/A).P(A)[/tex]

By substituting the values, we get

→                  [tex]= 0.79\times 0.05[/tex]

→                  [tex]= 0.0395[/tex]

(b)

The probability that a person does not have tuberculosis will be:

→ [tex]P(A') = 1-P(A)[/tex]

→            [tex]=1-0.05[/tex]

→            [tex]= 0.95[/tex]

(c)

The person does not have tuberculosis as well as have a (+) test, the probability will be:

→ [tex]P(A' \cap B) = P(A').P(B/A')[/tex]

→                  [tex]= 0.95\times 0.08[/tex]

→                  [tex]= 0.076[/tex]

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