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Mathematics Question on Probability

A laboratory blood test is 99%99\% defecting a certain disease when it is, infact present.However the test also yields a falls positive result for 0.5%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%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

The correct answer is: 22133\frac{22}{133}
Let E1E_1=The person selected is suffering from certain disease,E2E_2=The person selected is not sufferinf from certain disease and A=The doctor
diagnoses correctly
Now, P(E1)=0.1%=0.001,P(E2)=1−0.001=0.999,P(E_1)=0.1\%=0.001, P(E_2)=1-0.001=0.999,
P(A∣E1)=99%=99100=0.99,P(A∣E2)=0.005%P(A|E_1)=99\%=\frac{99}{100}=0.99, P(A|E_2)=0.005\%
Therefore,by Bayes'theorem,
P(E1∣A)=P(E1)P(A∣E1)P(E1)P(A∣E1)+P(E2)P(A∣E2)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)}
=0.01×0.990.001×0.99+0.999×0.005=\frac{0.01×0.99}{0.001×0.99+0.999×0.005}
=990990+4995=\frac{990}{990+4995}
=22133=\frac{22}{133}