Q. 345.0( 1 Vote )

# Let d1, d2, d3 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 the 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?

Given:

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.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses