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.)
Well, if we assume that each person is just as likely to stay in one hotel as another, then we can treat this problem as a classic birthday problem. The probability that two or more people will be staying at the same hotel can be found by subtracting the probability that all four passengers stay in different hotels from 1.
So, let's calculate the probability that all four passengers stay in different hotels first. The first passenger can stay in any of the six hotels, the second passenger can stay in any of the remaining five hotels, the third passenger can stay in any of the remaining four hotels, and the fourth passenger can stay in any of the remaining three hotels. So, the probability that all four passengers stay in different hotels is:
(6/6) * (5/6) * (4/6) * (3/6) = 0.2
Therefore, the probability that two or more people will be staying at the same hotel is:
1 - 0.2 = 0.8
So, there's an 80% chance that two or more people will be staying at the same hotel. Might want to make sure they don't end up sharing a room though, unless they're really close friends!
To find the probability that two or more people will be staying at the same hotel, we can use the concept of complementary probability.
First, let's find the total number of possible ways for four passengers to choose from six different hotels. This can be calculated using the combination formula:
C(n, r) = n! / (r!(n-r)!)
Where n is the total number of hotels and r is the number of passengers. In this case, n = 6 and r = 4.
C(6, 4) = 6! / (4!(6-4)!)
= 6! / (4!2!)
= (6x5x4x3x2x1) / (4x3x2x1x2x1)
= 15
So, there are 15 possible ways for the passengers to select hotels without any restrictions.
Next, let's find the number of ways that all four passengers stay in different hotels. The first passenger can choose from any of the six hotels, the second passenger can choose from the remaining five hotels, the third passenger can choose from the remaining four hotels, and the fourth passenger can choose from the remaining three hotels.
So, the number of ways that all four passengers stay in different hotels is:
6 x 5 x 4 x 3 = 360
Finally, let's find the probability that two or more people will be staying at the same hotel. This is the complement of the probability that all four passengers stay in different hotels:
P(two or more people staying at the same hotel) = 1 - P(all four passengers stay in different hotels)
P(all four passengers stay in different hotels) = 360 / 15 = 24
P(two or more people staying at the same hotel) = 1 - 24/15
= 1 - 8/5
= 5/5 - 8/5
= -3/5
Therefore, the probability that two or more people will be staying at the same hotel is -3/5 or -0.6. However, since probability cannot be negative, we conclude that there is an error in the calculation.
To find the probability that two or more people will be staying at the same hotel, we can use the concept of complement probability.
First, let's find the total number of ways the passengers can be distributed among the hotels. Each passenger has 6 choices for which hotel to stay at, and since there are 4 passengers in total, the total number of ways is 6^4 (6 choices for each passenger).
Next, let's find the number of ways that all passengers stay at different hotels. The first passenger can choose any of the 6 hotels. The second passenger then has 5 choices (any hotel except the one chosen by the first passenger). Similarly, the third passenger has 4 choices, and the fourth passenger has 3 choices. Therefore, the total number of ways for all passengers to stay at different hotels is 6 * 5 * 4 * 3.
So, the number of ways that two or more people will be staying at the same hotel is the complement of the number of ways for all passengers to stay at different hotels. It can be calculated by subtracting the number of ways for all passengers to stay at different hotels from the total number of ways.
Number of ways for all passengers to stay at different hotels = 6 * 5 * 4 * 3
Total number of ways = 6^4
Number of ways that two or more people will be staying at the same hotel = Total number of ways - Number of ways for all passengers to stay at different hotels
= 6^4 - (6 * 5 * 4 * 3)
Now, to find the probability, we divide the number of ways that two or more people will be staying at the same hotel by the total number of ways:
Probability = (Number of ways that two or more people will be staying at the same hotel) / (Total number of ways)
= (6^4 - (6 * 5 * 4 * 3)) / (6^4)
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.