# A laboratory blood test is 99% effective in detecting a certain disease when it is in fact,

Question:

A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (that is, 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?

Solution:

Let E1 and E2 be the respective events that a person has a disease and a person has no disease.

Since E1 and E2 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.

$P\left(E_{1}\right)=0.1 \%=\frac{0.1}{100}=0.001$

$\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)=\mathrm{P}$ (result is positive given the person has disease $)=99 \%=0.99 \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)=\mathrm{P}$ (result is positive given that the person has no disease $)=0.5 \%=0.005$ Probability that a person has a disease, given that his test result is positive, is given by

$P\left(E_{1} \mid A\right)$

By using Bayes’ theorem, we obtain

$P\left(E_{1} \mid A\right)=\frac{P\left(E_{1}\right) \cdot P\left(A \mid E_{1}\right)}{P\left(E_{1}\right) \cdot P\left(A \mid E_{1}\right)+P\left(E_{2}\right) \cdot P\left(A \mid E_{2}\right)}$

$=\frac{0.001 \times 0.99}{0.001 \times 0.99+0.999 \times 0.005}$

$=\frac{0.00099}{0.00099+0.004995}$

$=\frac{0.00099}{0.005985}$

$=\frac{990}{5985}$

$=\frac{110}{665}$

$=\frac{22}{133}$