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by thephoenix » Wed Jan 27, 2010 9:33 am
Prob. of 1st car: 3/3=1
Prob. of 2nd car: 2/3
Prob. of 3rd car: 1/3
1*2/3*1/3=2/9
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by ajith » Wed Jan 27, 2010 9:36 am
armaan700+ wrote:pls help to solve this
There are a total of 27 permutations to ride the cars (3*3*3)

Out of which 3! ways are favorable (ABC, BCA, CBA, CAB, BAC , ACB)

so the probability = 6/27 = 2/9
Always borrow money from a pessimist, he doesn't expect to be paid back.
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