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# A car dealership has 8 cars in stock each having a diferent

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21 Feb 2006, 16:09
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A car dealership has 8 cars in stock each having a diferent price. A customer wants to buy 3 cars. What is the probability that the 2 most expensive cars will be selected?
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21 Feb 2006, 17:54
= 6/8c3=6/56=3/28
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21 Feb 2006, 19:13
Since each has a different car, assume that the two most expensive has been selected. Now we need to select 1 car out of the remaining 6. - 6C1 = 6 ways

Total number of ways to pick 3 cars from 8 = 8C3 = 56

P = 6/56 = 3/28
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21 Feb 2006, 19:16
Total cases of selecting 3 cars = 8C3 = 56

Lets assume 7 and 8 number cars are most expensive. Then in how many combinations of 3 selected cars, 7 and 8 will be there = 6

(1,7,8), (2,7,8).... (6,7,8)

So prob = 6/56 = 3/28
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21 Feb 2006, 21:05
ps_dahiya wrote:
Total cases of selecting 3 cars = 8C3 = 56

Lets assume 7 and 8 number cars are most expensive. Then in how many combinations of 3 selected cars, 7 and 8 will be there = 6

(1,7,8), (2,7,8).... (6,7,8)

So prob = 6/56 = 3/28

thnks got it..
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21 Feb 2006, 21:05
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