Assume that a disease has a prevalence of 0.3% in the population. A company had developed a diagnostic for this disease that is 97% reliable (meaning that it detects 97% of true cases) but has a false positive rate of 2%.
a) If a person is tested positive by the diagnostic test, what are the odds that the person has the disease (odds of A = probability of A / probability of not A)?
b) What is the probability that the person does not have the disease?
Any help would be much appreciated, thank you!
Although there are straightforward formulas for this kind of thing, you can often get a more intuitive feel for the situation by just drawing up a sample population. Imagine you have $100000$ people (a hundred thousand). Since $0.3$ percent of the population has the disease, it affects $300$ people.
Of those $300$ people, $97$ percent, or $291$ people, will test positive. The other $9$ people with the disease will test negative.
Of the remaining $99700$ people, $2$ percent, or $1994$ people, will test positive. The other $92706$ people without the disease will test negative.
Thus, of the $291+1994 = 2285$ people who test positive, only $291$ of them ($291/2285 = 0.127$, or $12.7$ percent) will actually have the disease. The other $87.3$ percent will be false positives.
The formula, from Bayes's Rule, is
$$ P(B \mid A) = \frac{P(A \mid B)P(B)}{P(A \mid B)P(B)+P(A \mid \neg B)P(\neg B)} $$
where $A = \text{tests positive}$ and $B = \text{has the disease}$, and the $\neg$ sign means "not."