# a probability question using percentages

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.

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The meaning of, say, $17$% is $\frac{17}{100}$, so translation from percent language to ordinary number language is automatic: $17$% **is** $0.17$. –  André Nicolas Mar 1 '12 at 6:29
Yea got that :) ... another quick question which I thought I shouldnt post a whole new topic for : the statement : "Only one tenth of 1% of the individuals in a certain population have a particular disease (an incident rate of .001), does that mean only 0.001 percent are affected ? –  Ahoura Ghotbi Mar 1 '12 at 6:36
$1$% is $1/100$, which is $0.01$. One tenth of that is $0.001$. But what you quoted says that, it says the indidence rate is $0.001$. It is definitely not $0.001$ percent, which is $0.00001$. –  André Nicolas Mar 1 '12 at 6:43

You need to use the following formula, which is undoubtedly part of your course material: $$P(A\cup B)=P(A)+P(B)-P(A\cap B). \qquad(\ast)$$

At least informally, this is fairly easy to see by drawing a diagram. Draw two intersecting circles, and label them $A$ and $B$. The probability of $A\cup B$ is kind of the weight of $A\cup B$. If we add $P(A)$ and $P(B)$, we have counted the probability of $A\cap B$ twice, so we must subtract it.

Let $A$ be the event "has auto insurance" and let $B$ be the event "has home insurance."

For Question a), we want $P(A\cap B)$. Using the formula $(\ast)$, we find that $0.75=0.60+0.40-P(A\cap B)$, and therefore $P(A\cap B)=0.25$.

For Question b), you want $P(B|A)$. Here you use the fact that $$P(A\cap B)=P(B|A)P(A).$$ From a), you know $P(A\cap B)$. And you certainly know $P(A)$. Now you can find $P(B|A)$.

Question c) is answered very much like Question b).

Remark: Note that $A$ and $B$ are not independent. If they were, we would have $P(A\cap B)=P(A)P(B)$. But we saw that $P(A\cap B)=0.25$. Note that $P(A)P(B)=0.24$. Not equal!

However, interestingly enough, the two numbers $0.25$ and $0.24$ are quite close to each other. Informally, although $A$ and $B$ are not independent, they are fairly close to being independent.

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yea it is part of the course, just saw it. but how about the third statement, statement C? how would I include that in my calculations ?? it cant be there for no reason right? :S –  Ahoura Ghotbi Mar 1 '12 at 6:07
and also part a mentions the customers who have BOTH types, so is what I am doing correct ? –  Ahoura Ghotbi Mar 1 '12 at 6:09
Thanks man, are you by any chance located in Canada, and available for tutoring? –  Ahoura Ghotbi Mar 1 '12 at 6:17

Yes, $P(A) = 6/10$ and $P(B) = 4/10$, but no, $A$ and $B$ are not independent. The third piece of information is telling you $P(A \cup B) = 75/100$.

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could you elaborate a bit more on how I can account for P(A union B) in my calculations? –  Ahoura Ghotbi Mar 1 '12 at 5:54
See Andre's answer –  Robert Israel Mar 1 '12 at 17:20