1.

By examining the chest X-ray, the probability that a person is diagonased with TB when he is actually suffering from it, is 0.99. The probability that the doctor incorrectlydiagnoses aperson to be having TB, on the basis of X-ray reports is 0.001. In a certain city , 1 in 1000 persons suffers from TB. A person is selected at random and is diagoanl to have TB. What is the chance that he actually has TB?

Answer»

Solution :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`.


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