Answer:
Let \[{{E}_{1}}\] be the event that person is suffering from cancer and \[{{E}_{2}}\] be the event that person is not suffering from cancer. Let E be the event that the doctor diagnoses that person has cancer. . Then, \[P({{E}_{1}})=\frac{1}{1000}\] and \[P({{E}_{2}})=1-\frac{1}{1000}=\frac{999}{1000}\] \[\therefore \] \[P\left( \frac{E}{{{E}_{1}}} \right)=P\] (Cancer is detected when a person is actually suffering) \[=0.990=\frac{990}{1000}\] and \[P\left( \frac{E}{{{E}_{2}}} \right)=P\] (Cancer is detected when a person is not actually suffering) \[=0.001=\frac{1}{1000}\] P (selected person has actually cancer) \[=P\left( \frac{E}{{{E}_{1}}} \right)=\frac{P({{E}_{1}})\cdot P\left( \frac{E}{{{E}_{1}}} \right)}{\left[ P({{E}_{1}})\cdot P\left( \frac{E}{{{E}_{1}}} \right)+P({{E}_{2}})\cdot P\left( \frac{E}{{{E}_{2}}} \right) \right]}\] [by Baye's theorem] \[=\frac{\left( \frac{1}{1000} \right)\left( \frac{990}{1000} \right)}{\left( \frac{1}{100} \right)\left( \frac{990}{1000} \right)+\left( \frac{999}{1000} \right)\left( \frac{1}{1000} \right)}\] \[=\frac{\frac{990}{1000000}}{\left( \frac{990+999}{1000000} \right)}=\frac{990}{1989}=\frac{110}{221}\] Hence, the probability that a selected person has actually cancer, is \[\frac{110}{221}.\]
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