question_answer

                                   By examining the MRI result, the MRI results, the probability that cancer is detected when a person is actually suffering, is 0.99. The probability of a healthy person diagnosed to have cancer is 0.001. In a certain city, 1 in 1000 people suffers from cancer. A person is selected at random and is diagnosed to have cancer. What is the; probability that he actually has cancer?

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 \[\frac{110}{221}.\]