Let the event D be that a person has the disease.
[tex]P(D)=\frac{3}{100},P(D^{\prime})=\frac{97}{100}[/tex]Now Let T be positive test and T' be a negative test result.
[tex]P(T|D^{\prime})=\frac{87}{100},P(T|D)=\frac{13}{100},P(T^{\prime}|D)=\frac{2}{100},P(T^{\prime}|D^{\prime})=\frac{98}{100}[/tex]So the probability that the person has the disease given that the test is positive is given by:
[tex]\begin{gathered} P(D|T)=\frac{P(D)P(T|D)}{P(D)P(T|D)+P(D^{\prime})P(T|D^{\prime})} \\ =\frac{\frac{3}{100}\times\frac{13}{100}}{\frac{3}{100}\times\frac{13}{100}+\frac{97}{100}\times\frac{87}{100}} \\ =\frac{13}{2826} \\ \approx0.0046 \end{gathered}[/tex]So only 0.46% of the people who tested positive will have the disease.