A woman lives in a country where only 1 out of 1000 people has the virus. There is a test available that is positive 5% of the time when the patient does not have it, negative 1% of the time when the patient does have it, and otherwise correct. Recall that we computed that the woman’s chance of having the virus, conditional on a positive test, is less than 1.9%. (In Bayesian parlance, we call the initial, unconditional, probability the “prior” and the resulting conditional probability, after updating based on observations, the “posterior.”)
Let the conditional probability we computed (1.9%) serve the role as the new prior. Compute the new probability that she has the virus (new posterior) based on her getting a second positive test.
If the results were independent, the answer would be straightforward 0.019. However, I know that if we go sequentially, Probability( the person has virus | when the result declared is positive) will keep increasing; but I am unable to write the Bayesian formula in this case. Please help.
Let A be the event that she has the disease. Let B, be the event that she tests positive. $$\Pr(A|B)=\frac{\Pr(B|A)\Pr(A)}{\Pr(B|A)\Pr(A)+\Pr(B|A')\Pr(A')}$$
Notice that
$\Pr(A)=0.001$
$\Pr(A')=0.999$
$\Pr(B|A)=0.99$
$\Pr(B|A')=0.05$
If we want the probability she has the disease given she 'tests positive twice' ( event C).
$P(C|A)=(0.99)^{2}$