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Question
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?
Solution
Let E1: Event that a person has TB
E2: Event that a person does not have TB
And H: Event that the person is diagnosed to have TB.
∴ P(E1) = `1/1000` = 0.001
P(E2) = `1 - 1/1000 = 999/1000` = 0.999
`"P"("H"/"E"_1)` = 0.99
`"P"("H"/"E"_2)` = 0.001
∴ `"P"("E"_1/"H") = ("P"("E"_1)*"P"("H"/"E"_1))/("P"("E"_1)*"P"("H"/"E"_1) + "P"("E"_2)*"P"("H"/"E"_2))`
= `(0.001 xx 0.99)/(0.001 xx 0.99 + 0.999 xx 0.001)`
= `0.99/(00.99 + 0.999)`
= `0.990/(0.990 + 0.999)`
= `990/1989`
= `110/221`
Hence, the required probability is `110/221`.
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