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

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A car dealership has 8 cars in stock each having a diferent [#permalink] New post 21 Feb 2006, 17:09
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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Re: combo/prob [#permalink] New post 21 Feb 2006, 18:54
= 6/8c3=6/56=3/28
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 [#permalink] New post 21 Feb 2006, 20:13
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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 [#permalink] New post 21 Feb 2006, 20:16
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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 [#permalink] New post 21 Feb 2006, 22:05
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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  [#permalink] 21 Feb 2006, 22:05
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