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.