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: .possible outcomes.
How many ways can they exit on different floors?
There are: .ways.
Answer: .
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