MHB What is the probability of a customer only insuring one non-sports car?

[Jason123]

Need help with a probability problem. I have the answer from the answer key, I just don't know how to figure it out.An insurance company examines its pool of auto insurance customers and gathers the following information:1) All customers insure at least one car.

2) 70% of the customers insure more than one car.

3) 20% of customers insure a sports car.

4) Of those customers who insure more than one car, 15% insure a sports car.
Calculate the probability that a randomly selected customer insures exactly one car and that car is not a sports car.

 

[Greg]

Hello Jason123 and welcome to MHB! :D

Any thoughts on where to begin?

Also, we ask that users do not post duplicate topics, thanks. I've deleted your other thread.

 

[Jason123]

I know I need to use the formula, pr(AnB)/Pr(B). For this problem I believe that pr(AnB) = the probability that someone has only one car and the car is not a sports car. And then divide that by the probability that someone has only one car. but I can't seem to get those numbers. Answer key says its .205.

 

[HallsofIvy]

Here's how I would do this problem (I'm not big on memorizing formulas):
"An insurance company examines its pool of auto insurance customers and gathers the following information:"
Imagine 1000 customers.

"1) All customers insure at least one car.

2) 70% of the customers insure more than one car."
So 700 insure more than one car, 300 insure one car.

"3) 20% of customers insure a sports car."
So 200 insure a sports car.

"4) Of those customers who insure more than one car, 15% insure a sports car."
Of the 700 customers who insure more than one car, .15(700)= 105 insure a sports car. Since 200 customers insured a sports car, that means there are 200- 105= 95 customers who insure only one car and that is a sports car. Since 300 customers insure one car, 300- 95= 205 customers insure one car and that car is not a sports car.

"Calculate the probability that a randomly selected customer insures exactly one car and that car is not a sports car."
That's easy now. Out of 1000 customers, 205 of them insure one car which is not a sports car. The probability is 205/1000= 0.205 or 20.5%.

 

[MarkFL]

Consider the following Venn diagram:

View attachment 6082

We must have:

$$x+0.7+0.095=1$$

$$x=0.205$$