Question

A laboratory blood test is $99\%$ effective in detecting a certain disease when it 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 $0.1$ percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

Answer 1

Let $E_1$ and $E_2$ be the respective events that a person has a disease and a person has no disease

.
Since $E_1$ and $E_2$ are events complimentary to each other.
$\therefore P\left(E_1\right)+P\left(E_2\right)=1$
$\Rightarrow P\left(E_2\right)=1-P\left(E_1\right)=1-0.001=0.999$

Let $A$ be the event that the blood test result is positive.
$P\left(E_1\right)=0.1$% $=\frac{0.1}{100}=0.001$

$P\left(A|E_1\right)=P\left(resultispositivegiventhepersonhasdisease\right)=99$% $=0.99$
$P\left(A|E_2\right)=P\left(resultispositivegiventhatthepersonhasnodisease\right)=0.5$% $=0.005$

Probability that a person has a disease, given that his test result is positive, is given by $P\left(E_1|A\right)$.
By using Baye's theorem, we obtain

$P\left(E_1|A\right)=\frac{P\left(E_1\right)\cdot P\left(A|E_1\right)}{P\left(E_1\right)\cdot P\left(A|E_1\right)+P\left(E_2\right)\cdot \left(A|E_2\right)}$
$=\frac{0.001\times 0.99}{0.001\times 0.99+0.999\times 0.005}$
$=\frac{0.00099}{0.00099+0.004995}$
$=\frac{0.00099}{0.005985}$
$=\frac{990}{5985}$
$=\frac{110}{665}$
$=\frac{22}{133}=0.165$