One percent of a population suffer from a certain disease. A diagnostic test gives a positive indication 97% of the time when an individual has the disease, and a negative response 95% of the time when an individual doesn't have the disease.
(i) Draw tree diagram
(ii) A person selected randomly reacts positively to the test. What is the probability that this person actually has the disease.
(iii) Calculate the probability that the test yields correct diagnosis for an individual chosen at random
Here's my tree diagram:
For part (ii) I used Baye's rule and said:
$$\frac{P(D∩+)}{P(D∩+)+ P(no D∩+)}$$
For part (iii) I simply said:
$$P(D∩+)+P(no D∩-)$$
So far am I correct? Please advise.
At no point in this question does it say use Bayes' rule, and as you are asked to draw the contingency tree you are expected to derive the asked for results from that tree.
For part (ii) you are probably expected to take the ratio of the + leaf on the upper branch (the probability that you tested positive and have the disease) to the sum of the probabilities of the + leaves (the probability that you test positive); that is $\frac{0.01\times 0.97}{0.01\times 0.97+0.99\times 0.05}=0.16385$, or $\approx 16\%$
For part (iii) you are probably expected to sum the probabilities of the correct diagnosis leaves in the tree, that is the + leaf on the upper branch and the - leaf on the lower branch to get $0.01 \times 0.97+ 0.99\times 0.95=0.9502$ (or $\approx 95\%$)
Which should agree with the results of applying Bayes' rule since the tree is a visualisation of the rule.