Q. 475.0( 1 Vote )

By examining the

Answer :

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


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]

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