Question

(1 point) In a random sample of 1,000 people, it is found that 6.7% have a liver ailment. Of those who have a liver ailment, 9% are heavy drinkers, 50% are moderate drinkers, and 41% are nondrinkers. Of those who do not have a liver ailment, 10% are heavy drinkers, 40% are moderate drinkers, and 50% are nondrinkers. If a person is chosen at random, and he or she is a heavy drinker, what is the empirical probability of that person having a liver ailment?

Answer 1

By using Bayes' theorem:

P(Liver ailment | Heavy drinker)

10.09 + 0.067 0.00 +0.067 +0.10 *(1-0.067)

= 0.060707