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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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    $1$% 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 $0.00001$.2012-03-01

2 Answers 2

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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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    Thanks man, are you by any chance located in Canada, and available for tutoring?2012-03-01
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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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    See Andre's answer2012-03-01