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

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Discussion Overview

The discussion revolves around calculating the probability that a randomly selected customer insures exactly one non-sports car, based on provided statistics about an insurance company's customer base. The scope includes mathematical reasoning and probability theory.

Discussion Character

  • Mathematical reasoning
  • Homework-related
  • Exploratory

Main Points Raised

  • One participant presents the problem and outlines the relevant statistics from the insurance company regarding customer car insurance.
  • Another participant suggests using the formula pr(AnB)/Pr(B) to approach the problem but struggles to derive the necessary probabilities.
  • A different participant proposes a method involving a hypothetical sample of 1000 customers to simplify the calculations, breaking down the numbers based on the provided statistics.
  • This participant calculates that 205 customers insure one car that is not a sports car, leading to a derived probability of 0.205 or 20.5%.
  • A later reply introduces a Venn diagram representation to further clarify the relationships between the different groups of customers and confirms the calculated probability.

Areas of Agreement / Disagreement

There is no explicit consensus on the method of calculation, as participants present different approaches and interpretations of the problem. Some participants agree on the final probability derived, while others focus on the methodology and calculations leading to that result.

Contextual Notes

Participants express uncertainty regarding the initial steps and the application of probability formulas. There are also assumptions made about the distribution of customers that may not be explicitly stated.

Who May Find This Useful

This discussion may be useful for individuals interested in probability theory, particularly in the context of real-world applications such as insurance statistics and customer behavior analysis.

Jason123
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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.
 
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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.
 
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.
 
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%.
 
Consider the following Venn diagram:

View attachment 6082

We must have:

$$x+0.7+0.095=1$$

$$x=0.205$$
 

Attachments

  • insurancevenn.png
    insurancevenn.png
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