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?

Let events E1, E2, E3 be the following:

E1 be the event that person has TB and E2 be the event that the person does not have TB


Total persons = 1000


So,


Now, Let E be the event that the person is diagnosed to have TB


P(E|E1) is the probability that TB is detected when a person is actually suffering


P(E|E2) the probability of an healthy person diagnosed to have TB


So, P(E|E1) = 0.99 and P(E|E2) = 0.001


Now, we have to find the probability that the person actually has TB


We use Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.


∴


P(E1|E) is the probability that person actually has TB









[Divide by 9 both numerator and denominator]


47