Question

2% of the population have a certain blood disease of a serious form: 10% have it in a mild form; and 88% don’t have it at all. A new blood test is developed; the probability of testing positive is 9/10  if the subject has the serious form, 6/10 if the subject has the mild form, and 1/10  if the subject doesn’t have the disease. A subject is tested positive. What is the probability that the subject has serious form of the disease?

Answer 1

Let event A₁ : Disease in serious form,

event A₂ : Disease in mild form, 

event A₃ : Subject does not have disease, 

event B: Subject tests positive. 

P(A₁ ) = 0.02, P(A₂) = 0.1, P(A₃) = 0.88 

The probability of testing positive is 9/10 if the subject has the serious form, 6/10 if the subject has the mild form, and 1/10 if the subject doesn’t have the disease. 

∴ P(B/A₁) = 0.9, P(B/A₂) = 0.6, P(B/A₃) = 0.1 

P(B) = P(A₁) P(B/A₁) + P(A₂) P(B/A₂ ) + P(A₃) P(B/A₃)

= 0.02 × 0.9 + 0.1 × 0.6 + 0.88 × 0.1 

= 0.166 

Required probability = P(A₁ /B)

By Baye’s theorem