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:

enter image description here

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.

  • $\begingroup$ So far it is correct, you are almost done. What remains to be done is the calculation of the joint probabilities via the conditional probabilities you can read from the tree, e.g. $P(D \wedge +)=P(D)P(+\vert D)$,... $\endgroup$ May 23 '16 at 10:29
  • $\begingroup$ Apologies, is that for part (iii) the question doesn't seem to ask for that $\endgroup$
    – Modrisco
    May 23 '16 at 10:35
  • 1
    $\begingroup$ For (iii) you are asked to provide $P(+\vert D)p(D)+P( -\vert no\ D)p(no\ D)=P(+\wedge D)+ P(-\wedge no\ D)$, which is precisely what you wrote. "Individual chosen at random" means that you have to add the probablities of all branches of the diagram that yield a correct diagnosis. -- Why do you think that something else was meant by the question? Am I overlooking something? $\endgroup$ May 23 '16 at 10:46
  • $\begingroup$ Basically I thought I left something out, you didn't overlook anything, thanks for answering my question. $\endgroup$
    – Modrisco
    May 23 '16 at 10:51

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.


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