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

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

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

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

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

$=\frac{0.00099}{0.005985}$

$=\frac{990}{5985}$

$=\frac{110}{665}$

$=\frac{22}{133}$