There are three passengers on an airport shuttle bugs that makes stops at four different hotels. What is the probability that all three passengers are staying at different hotels?
If it is equally likely that each passenger can stay at any of the three hotels:
-- the first passenger can stay at any hotel ---> 4/4 = 1
-- the probability that the second passenger stays at the same hotel ---> 1/4
-- the probability that the third passenger stays at the same hotel ---> 1/4
If these are independent choices: the probability is 1 · ¼ · ¼ = 1/16.
[However, all three persons could be from the same family, which means that the choices are not independent.]