Question

An airport limo has 4 passengers and stops at 6 different hotels. What is the probability that two or more people will be staying at the same hotel?

Step 1

**Concept of probability:**

Probability deals with the likelihood of occurrence of a given event. The probability value lies between 0 and 1. An event with probability 1 is considered as certain event and an event with probability 0 is considered as an impossible event. The probability of 0.5 infers of having equal odds of occurring and not occurring of an event.

The general formula to obtain probability of an event A is,

*P*(*A*) = (number of favourable elements for event *A*)/(Total number of elements in the sample space).

The basic properties of probability are given below:

Step 2

**Find the probability that none of the passengers stay in the same hotel:**

The total number of hotels is 6.

The number of passengers is 4.

Here, the objective is to find the probability that 2 or more passengers will be staying in the same hotel.

To reach the objective, it is important to find the opposite probability. That is, the probability that every passenger stays in the different hotel (no two or more passengers stays in the same hotel).

Let the 4 passengers be denoted by *A*, *B*, *C *and *D*.

- There are 6 different hotels for the passenger
*A*to stay. That is,*n*(*A*) = 6. - Then, there are 5 different hotels for the passenger
*B*to stay other than the hotel of*A*. That is,*n*(*B*) = 5. - Next, there are 4 different hotels for the passenger
*C*to stay other than the hotels of*A*and*B*. That is,*n*(*C*) = 4. - After that, there are 3 different hotels for the passenger
*D*to stay other than the hotels of*A*,*B*and*C*. That is,*n*(*D*) = 3.

The total number of hotels is *n*(*S*) = 6.

The probability that none of the passengers stay in the same hotel is obtained as (5/18) from the calculation given below:

Step 3

**Find the probability that two or more passenger’s will be staying in the same hotel:**

The two or more passenger’...

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