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?

Answer 1

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`.