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# 7 cars compete in a race and finish at different times. find

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CEO
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7 cars compete in a race and finish at different times. find [#permalink]  14 Sep 2003, 12:31
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7 cars compete in a race and finish at different times. find the probability that prat who drives the blue car finishes ahead of martin who drives a red car?
GMAT Instructor
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Re: PS : Probability # 3 [#permalink]  14 Sep 2003, 19:47
praetorian123 wrote:
7 cars compete in a race and finish at different times. find the probability that prat who drives the blue car finishes ahead of martin who drives a red car?

Since there is no difference in the cars, each car will be equally likely to be ahead of the other.

1/2.
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
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MFE, Haas School of Business, UC Berkeley, Class of 2005
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Intern
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Re: PS : Probability # 3 [#permalink]  14 Sep 2003, 20:12
But I think the statement is about one particular car finishing ahead of another specific car from the group so, the number of events is 6 and the total number of classifications would be 7!, therefore the answer may be
6/7! = 1/840

Am I right?
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No, I'm not. I apologize, I got awfully confused.
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It should be 2/7*1/2 = 2/14 = 1/7

P of selecting 2 cars from 7 and P of each of the two being ahead of the other.
Senior Manager
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I agree with Akamaibrah, 1/2

From all the posible 7! outcomes of the race, one specific car will be ahead of other specific car the same number of times since all have the same prob.

Although I would assign a higher probability of winning to the other car since "martin" is driving.
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I cannot digest akamaibrah logic. I hope he will explain it.

I get 1/2

There 7 postions so there 7! ways
if you fix the postions of martin and prat then you have

5! * ( 6+5+4+3+2+1 ) = 21 * 5!

so P = 21 * 5! / 7! = 21/42 = 1/2
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I've seen this type of question before and the best explanation for 1/2 is out of all the possible scenarios, half the time one finishes ahead of the other and half the time it's the other way around. Dont know if that makes sense.
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