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# Let ${d}_{1},{d}_{2},{d}_{3}$ be three mutually exclusive diseases. Let S be the set of observable symptoms of these diseases. A doctor has the following information from a random sample of 5000 patients: 1800 had disease d1, 2100 has disease d2, and others had disease d3. 1500 patients with disease d1, 1200 patients with disease d2, and 900 patients with disease d3 showed the symptom. Which of the diseases is the patient most likely to have?

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Solution

## Events E1, E2, E3 and S be the events defined as follows: E1: The patient had disease d1 E2: The patient had disease d2 E3: The patient had disease d3 S: The patient showed the symptom E1, E2, and E3 are mutually exclusive and exhaustive events. $\mathrm{P}\left({E}_{1}\right)=\frac{1800}{5000}=\frac{18}{50}\phantom{\rule{0ex}{0ex}}\mathrm{P}\left({E}_{2}\right)=\frac{2100}{5000}=\frac{21}{50}\phantom{\rule{0ex}{0ex}}\mathrm{P}\left({E}_{3}\right)=\frac{1100}{5000}=\frac{11}{50}$ Now, $\mathrm{P}\left(\frac{S}{{E}_{1}}\right)$ = Probability that the patient showed symptom given that patient had disease d1 = $\frac{1500}{5000}=\frac{15}{50}$ $\mathrm{P}\left(\frac{S}{{E}_{2}}\right)$ = Probability that the patient showed symptom given that patient had disease d2 = $\frac{1200}{5000}=\frac{12}{50}$ $\mathrm{P}\left(\frac{S}{{E}_{3}}\right)$ = Probability that the patient showed symptom given that patient had disease d3 = $\frac{900}{5000}=\frac{9}{50}$ Using Bayes theorem, we have Probability that patient had disease d1 such that symptom of d1 showed = $\mathrm{P}\left(\frac{{E}_{1}}{S}\right)=\frac{\mathrm{P}\left({E}_{1}\right)\mathrm{P}\left(\frac{S}{{E}_{1}}\right)}{\mathrm{P}\left({E}_{1}\right)\mathrm{P}\left(\frac{S}{{E}_{1}}\right)+\mathrm{P}\left({E}_{2}\right)\mathrm{P}\left(\frac{S}{{E}_{\mathit{2}}}\right)+\mathrm{P}\left({E}_{3}\right)\mathrm{P}\left(\frac{S}{{E}_{\mathit{3}}}\right)}=\frac{\frac{18}{50}×\frac{15}{50}}{\frac{18}{50}×\frac{15}{50}+\frac{21}{50}×\frac{12}{50}+\frac{11}{50}×\frac{9}{50}}=\frac{270}{621}$ Probability that patient had disease d2 such that symptom of d2 showed = $\mathrm{P}\left(\frac{{E}_{2}}{S}\right)=\frac{\mathrm{P}\left({E}_{2}\right)\mathrm{P}\left(\frac{S}{{E}_{2}}\right)}{\mathrm{P}\left({E}_{1}\right)\mathrm{P}\left(\frac{S}{{E}_{1}}\right)+\mathrm{P}\left({E}_{2}\right)\mathrm{P}\left(\frac{S}{{E}_{\mathit{2}}}\right)+\mathrm{P}\left({E}_{3}\right)\mathrm{P}\left(\frac{S}{{E}_{\mathit{3}}}\right)}=\frac{\frac{21}{50}×\frac{12}{50}}{\frac{18}{50}×\frac{15}{50}+\frac{21}{50}×\frac{12}{50}+\frac{11}{50}×\frac{9}{50}}=\frac{252}{621}$ Probability that patient had disease d3 such that symptom of d3 showed = $\mathrm{P}\left(\frac{{E}_{3}}{S}\right)=\frac{\mathrm{P}\left({E}_{3}\right)\mathrm{P}\left(\frac{S}{{E}_{3}}\right)}{\mathrm{P}\left({E}_{1}\right)\mathrm{P}\left(\frac{S}{{E}_{1}}\right)+\mathrm{P}\left({E}_{2}\right)\mathrm{P}\left(\frac{S}{{E}_{\mathit{2}}}\right)+\mathrm{P}\left({E}_{3}\right)\mathrm{P}\left(\frac{S}{{E}_{\mathit{3}}}\right)}=\frac{\frac{11}{50}×\frac{9}{50}}{\frac{18}{50}×\frac{15}{50}+\frac{21}{50}×\frac{12}{50}+\frac{11}{50}×\frac{9}{50}}=\frac{99}{621}$ Thus, the patient is most likely to have the disease d1.

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