Thread: Probaility And Combination Question

Question:

7 passengers enter an elevator on the first floor of an 11-floor building.
Assume that the egress pattern to floors 2,3,....,11 is the same. What is the probability that all get off at different floors?

My Answer:

Number of objects to be arranged = 7passengers + (11floor - first floor - 1) = 16 objects
Possible arrangements = 16C7 = ways of egress
Ways of all get off at different floors = 9C2 ( 9 floors, 2 floors with no passenger gets out)

P(probability that all get off at different floors) = 9C2/16C7 = 0.00315 <---------Which is WRONG!!!!!

The correct answer is 0.0605, but I have no idea on how to do it, I am so frustrated ...

Hello, csy0961743!

7 passengers enter an elevator on the first floor of an 11-floor building.
Assume that the egress pattern to floors 2,3, ... ,11 is the same.
What is the probability that all get off at different floors?

Each of the 7 people have 10 choices of floors on which to exit.

. . There are: .$\displaystyle 10^7$ possible outcomes.


How many ways can they exit on different floors?

There are: .$\displaystyle _{10}P_7 \:=\:\dfrac{10!}{3!} \:=\:604,\!800$ ways.


Answer: .$\displaystyle \dfrac{604,\!800}{10,\!000,\!000} \;=\;0.06048$