Question

Let d₁, d_2, d₃ be three mutually exclusive diseases. Let S = [S₁,  S₂, ..., S₆] be the set of observable symptoms of these diseases. For example, S₁ is the shortness of breadth, S₂ is the loss of weight, S₃  3,500 with disease d₂ and, 3,300 with disease d₃. Also, 3,100 patients with disease d₁, 3,300 with disease d₂ and 3,000 with disease d₃. Show the symptoms S. Knowing that the patient has symptoms S, the doctor wishes to determine the patient's illness. On the basis of this informations what should the doctor conclude ?

Answer 1

Let D₁ denote the event that the patient has disease d₁. The events D₂ and D₃ are defined similarly.

Let S be the event that the patient shows the symptom S,

Using Bayes' theorem, we get

P(D₁/S)  =  The probability that the patient has disease d₁ knowing that he/she has symptoms S₁, S₂, ...., S₆

P(D₁/S)

Thus, knowing that the patient has symptoms S₁, S₂,...,S₃, the probability that he has disease d₁ is 0.33, the probability that he has disease d₃ is 0.32, the probability that he has disease d₃ is 0.32. Therefore, the doctor should conclude that the patient is most likely to have disease d₂.