question_answer

Let \[{{d}_{1}},\] \[{{d}_{2}}\] and \[{{d}_{3}}\] be three mutually exclusive diseases and \[S=({{s}_{1}},\,{{s}_{2}},\,{{s}_{3}},....,{{s}_{6}})\] be the set of observable symptoms of these disease. For example, \[{{s}_{1}}\] is the shortness of breath, \[{{s}_{2}}\] is loss of weight, \[{{s}_{3}}\] is fatigue, etc. Suppose a random sample of 10000 patients contains 3200 patients with disease \[{{d}_{1}},\] 3500 with disease \[{{d}_{2}}\] and 3000 with disease \[{{d}_{3}}\]  Also, 3100 patients with disease \[{{d}_{1}},\] 3300 with disease \[{{d}_{2}}\] and 3000 with disease \[{{d}_{3}}\] show the symptoms S. Knowing that the patient has symptom S, the doctor wishes to determine the patient's illness. On the basis of this information, what should the doctor conclude?

Answer:

Let \[{{E}_{1}}\] denote the event that the patient has disease \[{{d}_{1}};i=1,\,\,2,\,\,3\]and A be the event that the patient has symptom S. Then,    \[P\,({{E}_{1}})=\frac{3200}{10000}=\frac{32}{100},\] \[P\,({{E}_{2}})=\frac{3500}{10000}=\frac{35}{100}\] and       \[P\,({{E}_{3}})=\frac{3300}{10000}=\frac{33}{100}\] \[\therefore \]\[P\,(A\cap {{E}_{1}})=\frac{3100}{10000}=\frac{31}{100}\] \[P\,(A\cap {{E}_{2}})=\frac{3300}{10000}=\frac{33}{100}\] and \[P\,(A\cap {{E}_{3}})=\frac{3000}{10000}=\frac{30}{100}\] \[\therefore \]\[P\left( \frac{A}{{{E}_{1}}} \right)=\frac{P\,(A\cap {{E}_{1}})}{P\,({{E}_{1}})}=\frac{31/100}{32/100}=\frac{31}{32},\] \[P\left( \frac{A}{{{E}_{2}}} \right)=\frac{P\,(A\cap {{E}_{2}})}{P\,({{E}_{2}})}=\frac{33/100}{35/100}=\frac{33}{35}\] and \[P\left( \frac{A}{{{E}_{3}}} \right)=\frac{P\,(A\cap {{E}_{3}})}{P\,({{E}_{3}})}=\frac{30/100}{33/100}=\frac{30}{33}\] Using Baye?s theorem, \[P\left( \frac{{{E}_{1}}}{A} \right)=\frac{P\,({{E}_{1}})\cdot P\,(A/{{E}_{1}})}{P\,({{E}_{1}})\cdot P\,(A/{{E}_{1}})+P\,({{E}_{2}})\cdot P\,(A/{{E}_{2}})+P\,({{E}_{3}})\cdot P\,(A/{{E}_{3}})}\]\[=\frac{\frac{32}{100}\times \frac{31}{32}}{\frac{32}{100}\times \frac{31}{32}+\frac{35}{100}\times \frac{33}{35}+\frac{33}{100}\times \frac{30}{33}}=\frac{33}{94}\] \[P\left( \frac{{{E}_{2}}}{A} \right)=\frac{P\,({{E}_{2}})\cdot P\,(A/{{E}_{2}})}{P\,({{E}_{1}})\cdot P\,(A/{{E}_{1}})+P\,({{E}_{2}})\cdot P\,(A/{{E}_{2}})+P\,({{E}_{3}})\cdot P\,(A/{{E}_{3}})}\]\[=\frac{\frac{35}{100}\times \frac{33}{35}}{\frac{32}{100}\times \frac{31}{32}+\frac{35}{100}\times \frac{33}{35}+\frac{33}{100}\times \frac{30}{33}}=\frac{33}{94}\] and \[P\left( \frac{{{E}_{3}}}{A} \right)=\frac{P\,({{E}_{3}})\cdot P\,(A/{{E}_{3}})}{P\,({{E}_{1}})\cdot P\,(A/{{E}_{1}})+P\,({{E}_{2}})\cdot P\,(A/{{E}_{2}})+P\,({{E}_{3}})\cdot P\,(A/{{E}_{3}})}\]\[=\frac{\frac{33}{100}\times \frac{30}{33}}{\frac{32}{100}\times \frac{31}{32}+\frac{35}{100}\times \frac{33}{35}+\frac{33}{100}\times \frac{30}{33}}=\frac{33}{94}\] Clearly, \[P\,({{E}_{3}}/A)<P\,({{E}_{1}}/A)<P\,({{E}_{2}}/A)\]i.e. \[P\,({{E}_{2}}/A)\]is largest. Thus, the doctor should conclude that the patient is most likely to have discase d.