Question

By examining the chest X-ray, probability that TB is detected when a person is actually suffering is 0.99. The probability that the doctor diagnoses incorrectly that a person has T.B. on the basis of X-ray is 0.001. In a certain city 1 in 1000 persons suffers from T.B. A person is selected at random is diagnosed to have T.B. What is the chance that he actually has T.B.?

Answer 1

Let A, E₁ and E₂ denote the events that the doctor diagnoses correctly, the selected person suffers from TB and the selected person does not suffer from TB, respectively.

$$\begin{gathered} \therefore P\left(E_1\right)=\frac{1}{1000} \\ P\left(E_2\right)=\frac{999}{1000} \\ \mathrm{Now}, \\ P\left(A/E_1\right)=0.99 \\ P\left(A/E_2\right)=0.001 \\ \mathrm{Using}\mathrm{Bayes}\text{'}\mathrm{theorem},\mathrm{we}\mathrm{get} \\ \mathrm{Required}\mathrm{probability}=P\left(E_1/A\right)=\frac{P\left(E_1\right)P\left(A/E_1\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)} \\ =\frac{{\displaystyle \frac{1}{1000}}\times 0.99}{{\displaystyle \frac{1}{1000}}\times 0.99+{\displaystyle \frac{999}{1000}}\times 0.001} \\ =\frac{110}{221} \end{gathered}$$