Question

By examine the chest the chest $X-$ray probability that $T.B$ 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 bases of $X-$ray is $0.001$. In a certain city $1$ in $1000$ person suffers from $T.B$. A person is selected at random is diagnoses to have $T.B$. What is the chance that the actually has $T.B$?

Answer 1

Let $E_1=$The person selected is suffering from $T.B$

$E_2=$The person selected is not suffering from $T.B,A=$The doctor diagnoses correctly.

Then $P\left(E_1\right)=\frac{1}{1000},P\left(E_2\right)=\frac{999}{1000},P\left(A/E_1\right)=0.99$ and $P\left(A/E_2\right)=0.001$

Required probability $=P\left(E_1/A\right)=\frac{P\left(E_1\right)P\left(A/E_1\right)}{P\left(E\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)}$

$\frac{\frac{1}{1000}\times \frac{99}{100}}{\frac{1}{1000}\times \frac{99}{100}+\frac{999}{1000}\times \frac{1}{1000}}=\frac{99}{999+99}=\frac{11}{122}$