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_(1)`=Event that person has TB
`E_(2)`=Event that peron does not have TB
E=Event that the person is diagnosed to have TB
`thereforeP(E_(1))=1/1000=0.001,P(E_(2))=999/1000=0.999`
and `P(E//E_(1))=0.99and P(E//E_(2))=0.001`
`therefore P(E_(1)//E)=(P(E_(1)cdotP(E//E_(1))))/(P(E_(1))cdotP(E//E_(1))+P(E_(2))cdotP(E//E_(2)))`
`=(0.001xx0.99)/(0.001xx0.99+0.999xx0.001)`
`=(0.000990)/(0.000990+0.000999)` ltbr gt`=990/1989=110/221`