MHB Conditional Probability and Venn Diagrams

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The discussion revolves around 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. Participants clarify that "same make" means only selecting either two Chevrolets or two Jeeps, not a mix. The correct approach involves using combinations to determine the number of ways to select two Chevrolets from the total. There is confusion regarding when to apply the combination formula, particularly in this context. The key takeaway is understanding how to adjust the denominator to reflect only the relevant outcomes based on the condition provided.
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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