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 (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?
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 (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?

Solution:

Let \mathrm{E}_{1} represent a person who has a disease, \mathrm{E}_{2} represent a person who does not have a disease, and A represent a person who has a positive blood test.

E_{1} and E_{2} are events that are complementary to one another.

Then P\left(E_{1}\right)+P\left(E_{2}\right)=1

\Rightarrow P\left(E_{2}\right)=1-P\left(E_{1}\right)

Then \mathrm{P}\left(\mathrm{E}_{1}\right)=0.1 \%=0.1 / 100=0.001 and \mathrm{P}\left(\mathrm{E}_{2}\right)=1-0.001=0.999

Also P\left(A \mid E_{1}\right)=P( result is positive given that person has disease) =99 \%=0.99

And P\left(A \mid E_{2}\right)=P (result is positive given that person has no disease) =0.5 \%=0.005

Given a positive test result, the probability that a person has a disease is \mathrm{P}.

\left(\mathrm{E}_{1} \mid \mathrm{A}\right)

We have used Bayes’ theorem to arrive at our conclusion.

\mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)}

We acquire the values by substituting them.

=\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}

\Rightarrow \mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{A}\right)=\frac{22}{133}