# Conditional Probability and Venn Diagrams

#### navi

##### New member
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 #### Jameson

Staff member
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?

#### navi

##### New member
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)