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Re: 7 cars compete in a race and finish at different times. find [#permalink]
No, I'm not. I apologize, I got awfully confused.
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Re: 7 cars compete in a race and finish at different times. find [#permalink]
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.
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Re: 7 cars compete in a race and finish at different times. find [#permalink]
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. :wink:
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Re: 7 cars compete in a race and finish at different times. find [#permalink]
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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Re: 7 cars compete in a race and finish at different times. find [#permalink]
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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Re: 7 cars compete in a race and finish at different times. find [#permalink]
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