This question is confusing me as I am not used to seeing percentages in a possibility question.
in a large insurance agency - 60% of the customers have automobile insurance - 40% of the customers have homeowners insurance - 75% of the customers have on type or the other or both a) find the proportion of customers with both types of insurance. b) find the probability that a customer has homeowner insurance given that he has automobile insurance c) find the probability that a customer has automobile insurance given that he has home insurance
I am not asking for someone to solve this, I am just wondering if it would be the same logic as a normal "rolling the dice n times question"?
so far I have thought of this :
$P(A) = \frac{6}{10}$ customers have auto insurance
$P(B) = \frac{4}{10}$ have homeowner insurance
$P(C) = \frac{7.5}{10}$ have one type or the other or both
a) so we have to do $A \land B$ and since they are independent events we can just multiply $P(A) \times P(B)$ : $(6/10) * (4/10) = 24/100$
b) $P(B|A) = P(B \land A)/P(A) = (24/100)/6/10 = 4/10$
now thats as far as I have done but I started thinking that I did not use $C$ at all so I think I am doing something wrong and I should take $C$ into account but not sure how. could some one tell me if I am doing anything wrong.