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 d 2 and the others had disease d 3 . 1500 patients with disease d 1 , 1200 patients with disease d 2 and 900 patients with disease d 3 showed the symptom. Which of the diseases is the patient most likely to have?

Question

Philoid · Accepted Answer

Given:

1800 patients had disease d₁

2100 patients had disease d₂

1100 patients had disease d₃

Let us assume U₁, U₂, U₃ and A be the events as follows:

U₁ = Person with disease d₁

U₂ = Person with disease d₂

U₃ = Person with disease d₃

A = Showing the symptom S

Now,

⇒

⇒ P(A|U₁) = P(patient who has disease d₁ showed the symptom S)

⇒

⇒ P(A|U₂) = P(patient who has disease d₂ showed the symptom S)

⇒

⇒ P(A|U₃) = P(patient who has disease d₃ showed the symptom S)

⇒

Now we find

P(U₁|A) = P(The patient who showed symptom S has disease d₁)

P(U₂|A) = P(The patient who showed symptom S has disease d₂)

P(U₃|A) = P(The patient who showed symptom S has disease d₃)

Using Baye’s theorem:

⇒

Since the probability of P(U₁|A) is larger than other disease’s probability. The patient mostly likely to have the disease d₁.