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 E1 = Event that a person has TB
E2 = Event that a person does not have TB
And H = Event that the person is diagnosed to have TB.
So,
P(E1) = 1/1000 = 0.001, P(E2) = 1 – 1/1000 = 999/1000 = 0.999
P(H/E1) = 0.99, P(H/E2) = 0.001
Now, using Baye’s theorem we have
Therefore, the required probability is 110/221.
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