# Thread: Set Theory Application Problem

1. ## Set Theory Application Problem

An insurance agent offers his clients auto insurance, homeowner insurance, and renters insurance.
The purchase of homeowners insurance and renters insurance are mutually exclusive. The profile of agent's clients is as follows:

A= auto, B= homeowner, C= renter

i) 17 percent of these clients have none of these three products. -----> P(AuBuC)'= 17%
ii) 64 percent of clients have auto insurance -----------> P(A)=64 %
iii) Twice as many of the clients have homeowners insurance as have renters insurance -------------> P(C)= 2P(B)
iv) 35 percent of the clients have two of these products. -----------------> I am stuck on this one. I am thinking, P(A intersect B) + P(A intersect C) + P(B intersect C) = .35?
v) 11 percent of the clients have homeowners insurance, but not auto insurance. ------------------> P(B intersect A')

calculate percentage of clients who both have auto and renters insurance.

2. ## Re: Set Theory Application Problem Originally Posted by math951 An insurance agent offers his clients auto insurance, homeowner insurance, and renters insurance.
The purchase of homeowners insurance and renters insurance are mutually exclusive. The profile of agent's clients is as follows:

A= auto, B= homeowner, C= renter

i) 17 percent of these clients have none of these three products. -----> P(AuBuC)'= 17%
ii) 64 percent of clients have auto insurance -----------> P(A)=64 %
iii) Twice as many of the clients have homeowners insurance as have renters insurance -------------> P(C)= 2P(B)
iv) 35 percent of the clients have two of these products. -----------------> I am stuck on this one. I am thinking, P(A intersect B) + P(A intersect C) + P(B intersect C) = .35?
v) 11 percent of the clients have homeowners insurance, but not auto insurance. ------------------> P(B intersect A')
calculate percentage of clients who both have auto and renters insurance.
Note that $B~\&~C$ are mutually exclusive so $\mathcal{P}(B\cap C)=0$ , therefore $\mathcal{P}(A\cap B) +\mathcal{P}(A\cap C)=0.35$
"Twice as many of the clients have homeowners insurance as have renters insurance" means $\mathcal{P}(B)=2\mathcal{P}(C)$
$\mathcal{P}(B\cap A')=0.11$ moreover $\mathcal{P}(A'\cap B'\cap C')=0.17$

$\mathcal{P}(A\cup B\cup C)=\mathcal{P}(A)+\mathcal{P}(B)+\mathcal{P}(C)-\mathcal{P}(A\cap C)-\mathcal{P}(A\cap B)$

3. ## Re: Set Theory Application Problem

Wow this is fun. That makes sense of the mutually exclusive events because when I was reading the definition it said if two events are mutually exclusive they = empty set and I understood it but it is nice to see It now in use.

Also stupid question here but is there a difference between p(A intersect B intersect C)' and p(A' intersect B' intersect C').. I am assuming there is because of the way demorgans law works.

4. ## Re: Set Theory Application Problem

Also we can use Demorgan's law more than 2 events, right?

6. ## Re: Set Theory Application Problem Originally Posted by math951 Wow this is fun. That makes sense of the mutually exclusive events because when I was reading the definition it said if two events are mutually exclusive they = empty set and I understood it but it is nice to see It now in use.

Also stupid question here but is there a difference between p(A intersect B intersect C)' and p(A' intersect B' intersect C').. I am assuming there is because of the way demorgans law works.
Not a stupid question! I always tell my students that the only stupid question is the one they don't ask!
Here's a visual explanation. 