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.
$\endgroup$ 32 Answers
$\begingroup$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.
$\endgroup$ 3 $\begingroup$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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