There are y different travelers who each have a choice of vacationing at one of n different destinations. What is the probability that all y travelers will end up vacationing at the same destination?	Choose the correct choice from the following choices
A) 1/n!
B) n/n!
C) 1/n^y
D) 1/n^(y-1)
E) n/y^n
The simplest way to look at it is to assume that the first traveler picks a place. All the remaining travelers (given as y-1) now need to pick the same place for vacation.   As a result the probability is equivalent to (1/n) for each traveler to pick that same spot as the original traveler, which has to happen (y-1) times.   P=1n∗1n∗...∗1nP=1n∗1n∗...∗1n (y-1) times   P=1n(y−1)   The answer is 1/n^(y-1): 1/n^(y-1).
<result>D</result>
There are y different travelers who each have a choice of vacationing at one of n different destinations. What is the probability that all y travelers will end up vacationing at the same destination?