Question

On a Sunday night, there are four people on a bus. The bus has five stops left before it end sits route. Suppose each person will get off the bus at one of the stops, and will do so randomly
1- How many different ways could the people get off the bus?
2- What is the probability that all four people get off the bus on the first stop?
3- What is the probability that all four people get off the bus on the same stop?
4- What is the probability that exactly three of the four people get off the bus on the same stop?

Answer 1

There are two possible outcomes of this experiment either success p or failure q. It has a given number of trials and all trials are independent therefore it is binomial probability distribution.

1-  5 ways

2- 5/16

3- 1/16

4- 1/16

In the question given above n= 5 p =1/2 q= 1/2  r is the given point.

1. Part 1:

The number of ways in which different people get off the bus can be calculated using combinations since the order is not essential. Therefore

nCr= 5C4= 5 ways

2. Part 2:

The probability that all four people get off the bus on the first stop is given by :

P (x= 1)= 5C1 (1/2)^0(1/2)^4= 5(1/2)^4= 5/16

3. Part 3:- The probability that all four people get off the bus on the same stop.

P (x= x)= 5C5 (1/2)^0(1/2)^4= 1(1/2)^4= 1/16

4. Part 4- The probability that exactly three of the four people get off the bus on the same stop.

P (x= x)= 5C5 (1/2)^3(1/2)^1= 1(1/2)^4= 1/16

For binomial distribution click

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