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 & explanation
Correct answer: option 2
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 $P(E | E_1) = 0.99 \text{ and } P(E | E_2) = 0.001$
$∴ P(E_1 | E) = \frac{P(E_1) \cdot P(E | E_1)}{P(E_1) \cdot P(E | E_1) + P(E_2) \cdot P(E | E_2)}$
$= \frac{0.001 \times 0.99}{(0.001 \times 0.99) + (0.999 \times 0.001)}$
$= \frac{0.000990}{0.000990 + 0.000999} = \frac{990}{1989} = \frac{110}{221}$