By examining the chest X-ray, the probability that ${\rm{TB}}$ 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?

Let ${E_1} =$ Event that person has TB

${E_2} =$ Event that person does not have TB

$E =$ Event that the person is diagnosed to have TB
$\therefore$ $\quad P\left( {{E_1}} \right) = \frac{1}{{1000}} = 0.001,P\left( {{E_2}} \right) = \frac{{999}}{{1000}} = 0.999$

and $P\left( {E/{E_1}} \right) = 0.99$ and $P\left( {E/{E_2}} \right) = 0.001$

$\therefore$ $\quad P\left( {{E_1}/E} \right) = \frac{{P\left( {{E_1}} \right) \cdot P\left( {E/{E_1}} \right)}}{{P\left( {{E_1}} \right) \cdot P\left( {E/{E_1}} \right) + P\left( {{E_2}} \right) \cdot P\left( {E/{E_2}} \right)}}$

$= \frac{{0.001 \times 0.99}}{{0.001 \times 0.99 + 0.999 \times 0.001}}$

$= \frac{{0.000990}}{{0.000990 + 0.000999}}$

$= \frac{{990}}{{1989}} = \frac{{110}}{{221}}$