Conditional Probability and Venn Diagrams

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SUMMARY

The discussion centers on calculating the conditional probability of selecting two Chevrolets from a group of 8 Chevrolets and 4 Jeeps, given that both cars are of the same make. The correct interpretation of the problem requires understanding that the denominator must account only for the combinations of the same make. The appropriate combination formula, C(8,2), is used to determine the number of ways to select 2 Chevrolets from the total of 8. The confusion arises from the application of the combination formula in probability calculations.

PREREQUISITES
  • Understanding of conditional probability
  • Familiarity with combination formulas, specifically C(n, k)
  • Basic knowledge of Venn diagrams for visualizing probabilities
  • Ability to interpret probability statements correctly
NEXT STEPS
  • Study the concept of conditional probability in depth
  • Learn how to apply combination formulas in probability problems
  • Explore Venn diagrams and their applications in probability theory
  • Practice solving conditional probability exercises with varying scenarios
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Students studying probability theory, educators teaching statistics, and anyone looking to improve their understanding of conditional probability and combinatorial calculations.

navi
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I am having a hard time with the following exercise:

Assume for this problem that the company has 8 Chevrolets and 4 Jeeps, and two cars are selected randomly and given to sales representatives.
What is the probability of both cars being Chevrolets, given that both are of the same make?

I have tried many different things, but I do not even understand the question. I am assuming make refers to either a Jeep or a Chevrolet, so for each I am assigning them a Pr of 1/2. The union of both cars being chevrolets and being of the same make should be 2/8? I think this because it is a Chevrolet, and among Chevrolets there are 8 and you can only pick out 2...? I am totally lost :(
 
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navi said:
I am having a hard time with the following exercise:

Assume for this problem that the company has 8 Chevrolets and 4 Jeeps, and two cars are selected randomly and given to sales representatives.
What is the probability of both cars being Chevrolets, given that both are of the same make?

I have tried many different things, but I do not even understand the question. I am assuming make refers to either a Jeep or a Chevrolet, so for each I am assigning them a Pr of 1/2. The union of both cars being chevrolets and being of the same make should be 2/8? I think this because it is a Chevrolet, and among Chevrolets there are 8 and you can only pick out 2...? I am totally lost :(

Hi navi,

I agree this question is a little bite confusing. I interpret "given that both are the same make" as you can only show 2 Jeeps or 2 Chevrolet vehicles. You can't show 1 Jeep and 1 Chevrolet. This affects the denominator, where we divide by all possible outcomes.Let's focus on the Chevrolets. How many ways can we pick two 2 of them from the 8, assuming all of them are the same?
 
Jameson said:
Hi navi,

I agree this question is a little bite confusing. I interpret "given that both are the same make" as you can only show 2 Jeeps or 2 Chevrolet vehicles. You can't show 1 Jeep and 1 Chevrolet. This affects the denominator, where we divide by all possible outcomes.Let's focus on the Chevrolets. How many ways can we pick two 2 of them from the 8, assuming all of them are the same?

Could it be C(8,2)?

(sorry, I am confused as to when I should use the combination formula and when I shouldnt)
 

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