so I am struggling to construct a probability table using data from two trials that try to determine the effectiveness of a diagnostic test. I can generate a probability table for each trial separately, but I am required to construct a probability table for the diagnostic test as a whole.
The question is:
"Bodyworm is a disease that is deadly to adults but not children. Researchers have developed a test to determine if a patient has bodyworm or not. To determine the effectiveness of the test they conducted two trials. In the first trial, 25 known bodyworm patients were given the test, 20 of which returned a positive result. In the second trial, 100 subjects from the general population were chosen. Of these 100, 16 tested positive for bodyworm. Further investigation of these individuals revealed that in fact, only 8 of them had bodyworm."
Construct a probability table for the diagnostic test and hence find the prevalence of bodyworm.
One strategy I considered was combining the two samples into one big sample of 125, but that doesn't seem sensible. I was able to generate a probability table but I feel that it may be wrong. The first trial is a non-representative sample, the other is a representative one, surely you can't just sum them?
Please help! I am so stuck and confused.
Let $B$ be the event that a person has bloodworm. Let $P$ be the event that a person tests positive for bloodworm.
The first study tells us that the experimental probability that a person with bloodworm will test positive is $20/25$. Hence, \begin{align*} \Pr(P \mid B) & = \frac{20}{25} = \frac{4}{5}\\ \Pr(P^C \mid B) & = 1 - \frac{4}{5} = \frac{1}{5} \end{align*} The second study tell us that the experimental probability that somebody will test positive is $16/100$ and that the experimental probability that a person in the general population who tests positive has bloodworm is $8/16$. Hence, \begin{align*} \Pr(P) & = \frac{16}{100} = \frac{4}{25}\\ \Pr(P^C) & = 1 - \frac{4}{25} = \frac{21}{25}\\ \Pr(B \cap P) & = \frac{8}{100} = \frac{2}{25}\\ \Pr(B^C \cap P) & = \frac{8}{100} = \frac{2}{25}\\ \Pr(B \mid P) & = \frac{8}{16} = \frac{1}{2}\\ \Pr(B^C \mid P) & = \frac{8}{16} = \frac{1}{2} \end{align*} Since $$\Pr(B \cap P) = \Pr(P \mid B)\Pr(B)$$ we can solve for $\Pr(B)$.
\begin{align*} \Pr(B) & = \frac{\Pr(B \cap P)}{\Pr(P \mid B)}\\ & = \frac{\frac{2}{25}}{\frac{4}{5}}\\ & = \frac{2}{25} \cdot \frac{5}{4}\\ & = \frac{1}{10} \end{align*} Hence, $$\Pr(B^C) = 1 - \frac{1}{10} = \frac{9}{10}$$