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by anksgupta » Sun Jun 07, 2009 7:35 am
There are y different travelers who each have a choice of vacationing at one of n different detinations. What is the probability that all y travelers will end up vacationing at the same destination?

A. 1/n!

B. n/n!

C. 1/n^y

D. 1/n^(y-1)

E. n/y^n
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by vasbli » Sun Jun 07, 2009 7:44 am
Total # of choices: each of y travellers has n choices, hence n^y choices
Total # of successful choices: the group of y travellers can go to each of n destinations.

Answer: n/n^y=1/n^(y-1)

Choice: D
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by scoobydooby » Sun Jun 07, 2009 11:26 am
would go for C.

total possible arrangenments: n^y (each person n choices, y persons will have n^y choices )

only 1 way, all vacation together.

probability that all vacation together: 1/n^y

hence, C
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by vasbli » Sun Jun 07, 2009 12:41 pm
all vacation together is not 1 way, is 1 way x n locations.
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by scoobydooby » Sun Jun 07, 2009 9:14 pm
oops. thanks for pointing out vasbli !
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