Q. 325.0( 1 Vote )

# By examining the chest X - ray, probability that T.B is detected when a person is actually suffering is 0.99. The probability that the doctor diagnoses incorrectly that a person has T.B. on the basis of X - ray is 0.001. In a certain city 1 in 1000 persons suffers from T.B. A person is selected at random is diagnosed to have T.B. What is the chance that he actually has T.B.?

Let us assume U1, U2 and A be the events as follows:

U1 = Person having T.B

U2 = Person not having T.B

A = Diagnosing T.B disease

From the problem:

P(A|U1) = P(diagnosing T.B disease for the person who actually having T.B)

P(A|U2) = P(diagnosing T.B disease for the person who don’t actually have T.B)

Now we find

P(U1|A) = P(The person has T.B and diagnosed T.B)

Using Baye’s theorem:

The required probability is .

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