Let \[{{E}_{1}}\] = Event that a person has TB
\[{{E}_{2}}\] = Event that a person does not have TB
And H = Event that the person is diagnosed to have TB.
So,
\[P({{E}_{1}})\text{ }=\text{ }1/1000\text{ }=\text{ }0.001,\text{ }P({{E}_{2}})\text{ }=\text{ }1\text{ }\text{ }1/1000\text{ }=\text{ }999/1000\text{ }=\text{ }0.999\]
\[P(H/{{E}_{1}})\text{ }=\text{ }0.99,\text{ }P(H/{{E}_{2}})\text{ }=\text{ }0.001\]
Now, using Baye’s theorem we have
Therefore, the required probability is \[110/221\].