A screening test is carried out to detect a certain disease. It is found that 12% of the positive reports and 15% of the negative reports are incorrect. Assuming that the probability of a person getting positive report is 0.01, the probability that a person tested gets an incorrect report is

0.0027

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0.0173

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0.1497

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0.2100

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Solution

The correct option is C 0.1497
Let A = {person gets an incorrect report}

$E_{1} = {p e r s o n g e t t i n g + v e r e p o r t}$
$\Rightarrow P (E_{1}) = 0.01$
$E_{2} = {p e r s o n g e t t i n g - v e r e p o r t}$
$\Rightarrow P (E_{2}) = 0.99$

$P (A / E_{1}) = P (+ v e r e p o r t i s i n c o r r e c t)$
= 0.12

$P (A / E_{2}) = P (- v e r e p o r t i s i n c o r r e c t)$
= 0.15

By total probability
$P (A) = P (E_{1}) . P (A / E_{1}) + P (E_{2}) . (P A / E_{2})$
$= (0.01) (0.12) + (0.99) (0.15)$
$= 0.1497$

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