An airport limousine has four passengers and stops at six different hotels. What is the probability that two or more people will be staying at the same hotel? (Assume that each person is just as likely to stay in one hotel as another.)

Answer 1

This is 1 - probability that each person will stay in a different hotel.

There are 6!/2! ways to assign hotels to the four persons such that each person stays in a different hotel. The total number of ways to assign hotels to the person without any restriction is 6^4. The probability that each person will stay in a different hotel is thus:

6!/(2! 6^4) = 5/18

So, the probability that two or more will be staying in some hotel is 13/18.