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  • In a town,  of the population is HIV positive. A new diagnostic kit arrived in the market. This kit correc

In a town, 10 \%  of the population is HIV positive. A new diagnostic kit arrived in the market. This kit correctly identifies HIV positive individual 95 \% of the time and HIV negative individual 89 \% of time. A particular patient is tested by this kit and is found to be positive. Then the probability that the patient is actually positive________

Option: 1

0.4896


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Let        \mathrm{A= \{\text { Patient is found }+\mathrm{ve}\} }
            \mathrm{E_{1}= \{\mathrm{HIV}+\mathrm{ve}\}, E_{2}=\{\mathrm{HIV}-\mathrm{ve}\}}

\mathrm{\mathrm{P}\left(\mathrm{E}_{1}\right)= 10 \%=0.1\: \& \: \mathrm{P}\left(\mathrm{E}_{2}\right)=0.9}

\mathrm{\mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_{1}}\right)= \mathrm{P} \text { (kit gives correct result for +ve }\text { individual })=0.95}

\mathrm{\mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_{2}}\right)= \mathrm{P} \text { (kit gives wrong result for }-\mathrm{ve} \text { individual })=0.11}
 

Now using Baye's theorem
\mathrm{P(\text { Actually +ve }) =P\left(\frac{E_{1}}{A}\right)}
\mathrm{=\frac{P\left(E_{1}\right) \cdot P\left(\frac{A}{E_{1}}\right)}{P\left(E_{1}\right) \cdot P\left(\frac{A}{E_{1}}\right)+P\left(E_{2}\right) \cdot P\left(\frac{A}{E_{2}}\right)}}
\mathrm{=0.4896}

 

Posted by

Ritika Harsh

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