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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? |
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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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