Question

A doctor assumes that a patient has one of three diseases d₁, d₂ or d₃. Before any test, he assumes an equal probability for each disease. He carries out a test that will be positive with probability 0.8 if the patient has disease d₁, 0.6 if he has diseased d₂ and 0.4 if he has disease d₃. Given that the outcome of the test was positive, what is the probability that patient has disease d₃?

Answer 1

Answer:

2/9

Step-by-step explanation:

Hi,

Let d₁ denote the event that patient has disease d₁

P(d₁) = 1/3

Let d₂ denote the event that patient has disease d₁

P(d₂) = 1/3

Let d₃ denote the event that patient has disease d₁

P(d₃) = 1/3

Let T denote the event test result showing positive

Given that Test result showing positive if the patient has disease d₁

was 0.8,

So given P(T/d₁) = 0.8

Given that Test result showing positive if the patient has disease d₂

was 0.6,

So, given P(T/d₂) = 0.6

Given that Test result showing positive if the patient has disease d₃

was 0.4,

So, given P(T/d₃) = 0.4

Given that test result is positive,

P(T) = P(T ∩ d₁) + P(T ∩ d₂) + P( T ∩ d₃)

= P(d₁)P(T/d₁) + P(d₂)P(T/d₂) + P(d₃)P(T/d₃)

= 1/3*0.8 + 1/3*0.6 + 1/3*0.4

= 0.6

Required to find that patient has disease d₃ if the outcome of test

is positive = P(d₃/T) = P(d₃∩T)/P(T)

= P(d₃)P(T/d₃)/P(T)

= 1/3*0.6/0.4

= 2/9

Hope, it helps !