Question

A laboratory blood test is $99\%$ effective in detecting a certain disease when patient is infected actually by the disease. However, the test also yields a false positive result for $0.5\%$ of the healthy person tested 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?

Answer 1

Let

A: Person has the disease

B: Person does not have the disease

C: Test result is positive

$P\left(A|C\right)=\frac{P\left(A\right).P\left(C|A\right)}{P\left(B\right).P\left(C|B\right)+P\left(A\right).P\left(C|A\right)}$

$P\left(A\right)=$ Probability that person has disease $=0.1\%=\frac{0.1}{100}=0.001$

$P\left(C|A\right)=$ Probability that test result is positive, if the person has the disease $=99\%=\frac{99}{100}=0.99$

$P\left(B\right)=$ Probability that person does not have the disease $=99.9\%=\frac{99.9}{100}=0.999$

$P\left(C|B\right)=$ Probability that test result is positive, if the person does not have the disease $=0.005$

$P\left(A|C\right)=\frac{0.001\times 0.99}{0.999\times 0.005+0.001\times 0.99}$

$=\frac{99\times {10}^{-5}}{{10}^{-5}(499.5+99)}$

$=\frac{99}{598.5}$

$=\frac{990}{5985}$

$=\frac{22}{133}$

Therefore, required probability is $\frac{22}{133}$.