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?

By examining the. chest X-ray, the probability that TB is detected whe

1.	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?
Answer» 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`

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