Expert: Sherry Wallin - 11/18/2010

Question
the probability that a person selected at random from a population will exhibit the classic symptom of a certain disease is .2 and that a person selected at random has the disease is .23.The probability that a person who has the symptom also has the disease is .18.A person selected at random does not have the symptom.What is the probability that the person has the disease?

Answer
Jaspreet~

Let A be the event that a person demonstrates the classic symptom
Let B be the event that a person has the disease
Then A intersect B is the event that a person has the symptom and has the disease
Then A' is the event that a person does not demonstrate the classic symptom

We want the probability that B occurs given A' occurred, i.e., we want to know that if a person who does not demonstrate the symptom will have the disease.

You are given P(A) = .2
P(B) = .23
P(A intersects B) = .18
P(A') = 1 - .2 = .8

We want P(B/A') = P(A' intersects B)/(A')

You can draw a Venn Diagram to see that the intersection of A and B is .18 and if P(A) = .2 then the 'rest' of A is .02, .2 -.18 = .02
and
P(B) = .23 then the rest of B is .05, .23 - .18 = .05
A' intersects B at B minus the intersection of A and B, thus it is the 'rest' of B, i.e., .05

Therefore P(B/A')= P(A' intersects B)/(A') = .05/.8 = .0625

Math Prof