Question

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 d_1, 2100 has disease d₂, and others had disease d₃. 1500 patients with disease d_1, 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?

Answer 1

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.

$$\begin{gathered} \mathrm{P}\left(E_1\right)=\frac{1800}{5000}=\frac{18}{50} \\ \mathrm{P}\left(E_2\right)=\frac{2100}{5000}=\frac{21}{50} \\ \mathrm{P}\left(E_3\right)=\frac{1100}{5000}=\frac{11}{50} \end{gathered}$$
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 = $\mathrm{P}\left(\frac{E_1}{S}\right)=\frac{\mathrm{P}\left(E_1\right)\mathrm{P}\left({\displaystyle \frac{S}{E_1}}\right)}{\mathrm{P}\left(E_1\right)\mathrm{P}\left({\displaystyle \frac{S}{E_1}}\right)+\mathrm{P}\left(E_2\right)\mathrm{P}\left({\displaystyle \frac{S}{E_{\mathit{2}}}}\right)+\mathrm{P}\left(E_3\right)\mathrm{P}\left({\displaystyle \frac{S}{E_{\mathit{3}}}}\right)}=\frac{{\displaystyle \frac{18}{50}}\times {\displaystyle \frac{15}{50}}}{{\displaystyle \frac{18}{50}}\times {\displaystyle \frac{15}{50}}+{\displaystyle \frac{21}{50}}\times {\displaystyle \frac{12}{50}}+{\displaystyle \frac{11}{50}}\times {\displaystyle \frac{9}{50}}}=\frac{270}{621}$

Probability that patient had disease d₂ such that symptom of d₂ showed = $\mathrm{P}\left(\frac{E_2}{S}\right)=\frac{\mathrm{P}\left(E_2\right)\mathrm{P}\left({\displaystyle \frac{S}{E_2}}\right)}{\mathrm{P}\left(E_1\right)\mathrm{P}\left({\displaystyle \frac{S}{E_1}}\right)+\mathrm{P}\left(E_2\right)\mathrm{P}\left({\displaystyle \frac{S}{E_{\mathit{2}}}}\right)+\mathrm{P}\left(E_3\right)\mathrm{P}\left({\displaystyle \frac{S}{E_{\mathit{3}}}}\right)}=\frac{{\displaystyle \frac{21}{50}}\times {\displaystyle \frac{12}{50}}}{{\displaystyle \frac{18}{50}}\times {\displaystyle \frac{15}{50}}+{\displaystyle \frac{21}{50}}\times {\displaystyle \frac{12}{50}}+{\displaystyle \frac{11}{50}}\times {\displaystyle \frac{9}{50}}}=\frac{252}{621}$

Probability that patient had disease d₃ such that symptom of d₃ showed = $\mathrm{P}\left(\frac{E_3}{S}\right)=\frac{\mathrm{P}\left(E_3\right)\mathrm{P}\left({\displaystyle \frac{S}{E_3}}\right)}{\mathrm{P}\left(E_1\right)\mathrm{P}\left({\displaystyle \frac{S}{E_1}}\right)+\mathrm{P}\left(E_2\right)\mathrm{P}\left({\displaystyle \frac{S}{E_{\mathit{2}}}}\right)+\mathrm{P}\left(E_3\right)\mathrm{P}\left({\displaystyle \frac{S}{E_{\mathit{3}}}}\right)}=\frac{{\displaystyle \frac{11}{50}}\times {\displaystyle \frac{9}{50}}}{{\displaystyle \frac{18}{50}}\times {\displaystyle \frac{15}{50}}+{\displaystyle \frac{21}{50}}\times {\displaystyle \frac{12}{50}}+{\displaystyle \frac{11}{50}}\times {\displaystyle \frac{9}{50}}}=\frac{99}{621}$

Thus, the patient is most likely to have the disease d₁.