A laboratory blood test is 99 % effective in detecting a certain … - Vidyadip

Question

A laboratory blood test is $99\%$ effective in detecting a certain disease when its infection is present. However, the test also yields a false positive result for $0.5\%$ of the healthy person tested (i.e. if a healthy person is tested, then, with probability $0.005,$ the test will imply he has the disease). If $0.1\%$ of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

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Answer

Let $E_1 $ and $E_2 $ denote the events that a person has a disease and a person has no disease, respectively.
$E_1 $ and $E_2 $ are complimentary to each other.
$\therefore P(E_1) + P(E_2) = 1$
$\Rightarrow P(E_2) = 1 - P(E_1) = 1 - 0.001 = 0999$
let A denote the event that the blood test result is positive.
$\therefore P(E_1) = 0.1\% = 0.001$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=90\%=0.99$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\big)=0.5\%=0.005$
Using Baye's theorem, we get
Required probability $\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{0.001\times0.99}{0.001\times0.99+0.999\times0.005}$
$=\frac{990}{5985}=\frac{22}{133}$

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