By examining the chest X-ray, probability that T.B is detected when a person is actually suffering is 0.99. the probability that the doctor diagnoses incorrectly that a person has T.B. on the basis of X-ray is 0.001. in a certain city 1 in 100 persons suffers from T.B. A person is selected at random is diagnosed to have T.B. What is the chance that he actually has T.B.?

Explanation

Let E= event that the doctor diagonoses TB,
`E_1` =event that the person selected is suffering from TB, and
`E_2` =event that the person selected is not suffering from TB.
Then , `P(E_1)=1/1000andP(E_2)=(1-1/1000)=999/1000`.
`P(E//E-1)`= probability that TB is diagnosed, when the person actually has TB
`=99/100`
`P(E//E-2)`= probability that TB is diagnosed, when the person has no TB
`=1/1000`
Using Bayes's theorem, we have
`P(E_1//E)`= probability of a person actually having TB, if it is knows that he is diagonal to have TB
`(P(E//E_1) .P(E_1))/(P(E//E_1).P(E_1)+P(E//E_2).P(E_2))`
`=((99/100xx1/1000))/((99/100xx1/1000)+(1/1000xx999/1000))=110/221`.
Hence, the required probability is `110/221`.