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$10$% of the population in a town is $\text{HIV}\large ^{+}$. A new diagnostic kit for $\text{HIV}$ detection is available; this kit correctly identifies $\text{HIV}\large ^{+}$ individuals $95$% of the time, and $\text{HIV}\large ^{-}$ individuals $89$% of the time. A particular patient is tested using this kit and is found to be positive. The probability that the individual is actually positive is ______.
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If you dont like Bayes theorem then use this TREE METHOD:

Probability$=\dfrac{0.10\times 0.95}{0.10\times 0.95 + 0.90\times 0.11}$

$\hspace{1.9 cm} =0.4896$

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HIV10% of population

then HIV- is 90% of population

the kit correctly identifies HIV+ in 95% of time

the kit incorrectly identifies HIV- in (100-89)%= 11% of time

So, applying Baye's theorem 

        0.10⨉ 0.95                

.10⨉0.95 +  0.90⨉0.11

=0.4896

Answer:

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