Answer :

Let events E_{1}, E_{2}, E_{3} be the following:

E_{1} be the event that person has TB and E_{2} 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|E_{1}) is the probability that TB is detected when a person is actually suffering

P(E|E_{2}) the probability of an healthy person diagnosed to have TB

So, P(E|E_{1}) = 0.99 and P(E|E_{2}) = 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(E_{1}|E) is the probability that person actually has TB

[Divide by 9 both numerator and denominator]

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