empirical probability problem?
In a random sample of 1,000 people, it is found that 7.1% have a liver ailment. Of those who have a liver ailment, 6% are heavy drinkers, 50% are moderate drinkers, and 44% are nondrinkers. Of those who do not have a liver ailment, 12% are heavy drinkers, 49% are moderate drinkers, and 39% 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?
Comments
P[diseased & heavy drinker] = 7.1% *6% = 4.26%
P[healthy & heavy drinker ] = 12% * 92.1% = 11.052%
P[diseased | heavy drinker ] = 4.26/(4.26+11.052) = .2782 <-----