Question

a) A survey of the adults in a town shows that 8% have liver problems. Of these, it is also 

found that 25% are heavy drinkers, 35% are social drinkers and 40% are non-drinkers. Of 

those that did not suffer from liver problems, 5% are heavy drinkers, 65% are social 

drinkers and 30% do not drink at all. An adult is chosen at random, what is the probability 

that this person

i. Has a liver problems? (3 Marks)

ii. Is a heavy drinker (2 Marks)

iii. If a person is found to be a heavy drinker, what is the probability that this person 

has liver problem? (2 Marks)

iv. If a person is found to have liver problems, what is the probability that this person 

is a heavy drinker? (2 Marks)

v. If a person is found to be a non –drinker, what is the probability that this person has 

liver problems. (

Answer 1

Let $L$ be the event "An adult has a liver problems ", $H$ be the event "An adult is a heavy drinker", $S$ be the event "An adult is a soial drinker".

Given

$$P(L)=0.08, P(H|L)=0.25, P(S|L)=0.35,$$

$$P(H|L^C)=0.05, P(S|L^C)=0.65$$

$$P((H\cup S)^C|L)=0.4, P((H\cup S)^C|L^C)=0.3$$

i.

$P(L)=0.08$

ii.

$$P(H)=P(H|L)P(L)+P(H|L^C)P(L^C)=$$ $$=0.25(0.08)+0.05(1-0.08)=0.066$$

iii.

$$P(L|H)=\dfrac{P(H|L)P(L)}{P(H|L)P(L)+P(H|L^C)P(L^C)}=$$

$$=\dfrac{0.25(0.08)}{0.25(0.08)+0.05(1-0.08)}=\dfrac{0.02}{0.066}=\dfrac{10}{33}\approx$$

$$\approx0.3030$$

iv.

$$P(H|L)=0.25$$

v.

$$P(L|(H\cup S)^C)=$$

$$=\dfrac{P((H\cup S)^C|L)P(L)}{P((H\cup S)^C|L)P(L)+P((H\cup S)^C|L^C)P(L^C)}=$$

$$=\dfrac{0.4(0.08)}{0.4(0.08)+0.3(1-0.08)}=\dfrac{0.032}{0.308}=\dfrac{8}{77}\approx$$

$$\approx0.1039$$