Question

A laboratory blood test is 99% effective in detecting a certain diesease when it is fact 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 diease given that his test result is positive?

Answer 1

Let $E_1$ : the event that the person has disease and $E_2$ : the events that the person is healthy.

Then, $E_1$ and $E_2$ are mutually exclusive and exhaustive. .

Moreover, P $E_1$=0.1% = $\frac{0.1}{100}$=0.001 and P $E_2$=1-0.001=0.999

Let E: the event that test is positive,

$P\left(\frac{E}{E_1}\right)$= P(result is positive given the person has disease)

=99% $\frac{99}{100}$ =0.99

Probability that a person does not have disease and test result is positive.

$\therefore P\left(\frac{E}{E_2}\right)=0.5$

By using Baye's theorem, we obtain

$P\left(\frac{E_1}{E}\right)=\frac{P\left(\frac{E}{E_1}\right)P\left(E_1\right)}{P\left(\frac{E}{E_1}\right)P\left(E_1\right)+P\left(\frac{E}{E_2}\right)P\left(E_2\right)}$

$=\frac{0.001\times 0.99}{0.001\times 0.99+0.999\times 0.005}=\frac{0.000099}{0.00099+0.004995}=\frac{0.00099}{0.005985}=\frac{990}{5985}=\frac{110}{665}=\frac{22}{133}$