Your first stage had two branches, I labeled mine D and ND (disease and not disease)
Each of those has two branches which I labeled P and N (positive and negative)
The D branch is .005 and the ND is .995
The D-P branch is .9 and the D-N is .1
the ND-P is .02 and the ND-N is .98
So now you can find the prob of any combination, e.g.
"Find the probability that the person has the disease and the test is positive"
-----> that would be the path D,D-P
= (.005)(.9) = .0045
A certain disease is present in 1 in 200 of the population.In a mass screening program quick test for the disease is used, but the test is not totally reliable. For someone who does have the disease there is a probability of 0.9 that the test will prove positive, whereas for someone who does not have the disease there is a probability of 0.02 that the test will prove positive.
i.one person is selected at random and tested.
a.Draw a tree diagram, which illustrates one application of test.
b.Find the probability that the person has the disease and the test is positive.
c. Find the probability that the test is negative.
d.Given that the test is positive,find the probability that the person has the disease.
2 answers
do no. d
2 answers