Question

A laboratory blood test is \(99\%\) defecting a certain disease when it is, infact present.However the test also yields a falls 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 diesease).If \(0.1\%\) 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

The correct answer is: \(\frac{22}{133}\)
Let \(E_1\)=The person selected is suffering from certain disease,\(E_2\)=The person selected is not sufferinf from certain disease and A=The doctor
diagnoses correctly
Now, \(P(E_1)=0.1\%=0.001, P(E_2)=1-0.001=0.999,\)
\(P(A|E_1)=99\%=\frac{99}{100}=0.99, P(A|E_2)=0.005\%\)
Therefore,by Bayes'theorem,
\(P(E_1|A)=\frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}\)
\(=\frac{0.01×0.99}{0.001×0.99+0.999×0.005}\)
\(=\frac{990}{990+4995}\)
\(=\frac{22}{133}\)