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?
+23246 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.]
