Question

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, 6% are heavy drinkers, 50% are moderate drinkers, and 44% are nondrinkers. Of those who do not have a liver ailment, 10% are heavy drinkers, 45% are moderate drinkers, and 45% are nondrinkers.

If a person is chosen at random, and he or she is a heavy drinker, what is the probability of that person having a liver ailment?

Answer 1

Based on the given information:

Category	Nos	Heavery drinkers	moderate drinkers	Non drinkers
Liver ailment	67	6%	50%	44%
No live ailment	933	10%	45%	45%

Category	Nos	Heavery drinkers	moderate drinkers	Non drinkers	total
Liver aliment	67	4.02	33.5	29.5	67.0
No live aliment	933	93.3	419.9	419.9	933.0
total	1000	97.32	453.35	449.33	1000

P(has liver ailment / heavy drinker) =  P(has liver ailment and heavy drinker) / P (heavy drinker)

P(has liver ailment and  heavy drinker) = 4.02/ 1000

P (heavy drinker) = 97.32/ 1000

P(has liver ailment / heavy drinker) = ( 4.02/ 1000 ) / (97.32/ 1000) = 4.02 / 97.32= 0.0413

ANS : P(has liver ailment / heavy drinker) = 0.0413