By examining the. chest X-ray, the probability that TB is detected when a person is actually suffering is 0.99. The probability of an healthy person diagnosed to have TB is 0.001. In a certain city, 1 in 1000 people suffers from TB. A person is selected at random and is diagnosed to have TB. What is the probability that he actually has TB?
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`