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.
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