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?

$\frac{110}{221}$

The correct answer is Option (2) → $\frac{110}{221}$ ##

Let $E_1 = \text{Event that person has TB},$

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

and $E = \text{Event that the person is diagnosed to have TB}$

$∴P(E_1) = \frac{1}{1000} = 0.001, P(E_2) = 1 - P(E_1) = \frac{999}{1000} = 0.999$

and $P(E | E_1) = 0.99 \text{ and } P(E | E_2) = 0.001$

$∴ P(E_1 | E) = \frac{P(E_1) \cdot P(E | E_1)}{P(E_1) \cdot P(E | E_1) + P(E_2) \cdot P(E | E_2)}$

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

$= \frac{0.000990}{0.000990 + 0.000999} = \frac{990}{1989} = \frac{110}{221}$