Tests for tuberculosis like all other diagnostic tests are not perfect. QFT-G is one of such tests for tuberculosis. Suppose that for the population of adults that is taking the test, 5% have tuberculosis. The test correctly identifies 74.6% of the time adults with a tuberculosis and correctly identifies those without tuberculosis 76.53% of the time. Suppose that POS stands for the test gives a positive result and S means that the adult really has tuberculosis. What is the probability of an adult getting a POS result and truly NOT having tuberculosis?

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Answer:

The probability of having a positive test on a non-sick person is 24.32%.

Step-by-step explanation:

The probability of having a POS and not having tuberculosis, can be defined as the probability of getting a false positive.

The amount of positive test will be

1) The positive test of sick persons

[tex]P(test=pos;person=sick)=0.744*0.05=0.0372[/tex]

2) The positive test of not-sick persons (false positives)

[tex]P(test=pos;person=sick)=(1-0.744)*0.95=0.256*0.95=0.2432[/tex]

The amount of positive test will be

1) The negative test of non-sick persons

[tex]P(test=neg;person=non-sick)=0.7653*0.95=0.727[/tex]

2) The negative test of sick persons (false negatives)

[tex]P(test=neg;person=non-sick)=(1-0.7653)*0.05=0.2347*0.05=0.0117[/tex]

The probability of having a positive test on a non-sick person is 0.24 = 24.32%.

Suppose that,

The population of adults that is taking the test, 5% have tuberculosis.

The test correctly identifies 74.6% of the time adults with a tuberculosis and correctly identifies those without tuberculosis 76.53% of the time.

Suppose that, POS stands for the test gives a positive result and S means that the adult really has tuberculosis.

We have to find,

What is the probability of an adult getting a POS result and truly NOT having tuberculosis.

According to the question,

The probability of having a POS and not having tuberculosis, can be defined as the probability of getting a false positive.

  • The amount of positive test will be,

The positive test of sick persons,

[tex]p (test; positive, person=sick) = (0.744).(0.05) = 0.372[/tex].

The positive test of non sick persons (false positives),

[tex]p (test; positive, person=sick) = (1-0.744)(0.95) = 0.24[/tex]

  • The amount of positive test will be,

The negative test of non-sick persons,

[tex]p (test; negative, person=non sick) = (0.7653).(0.95) = 0.727[/tex]

The negative test of sick persons (false negatives),

[tex]p (test; positive, person=sick) = (1-0.7653)(0.05) = 0.0117[/tex]

Hence, The probability of having a positive test on a non-sick person is 0.24 = 24.32%.

For more information about Binomial distribution click the link given below.

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