the question goes like this:
"In a building of 10 floors, 5 people get inside an elevator on the entry floor. every one of them push independently a button of one of the 10 floors. what's the probability that the elevator would get exactly to the 5th floor and no higher?"
I thought of two ways to solve it, and each one gets me another result. so probably I do something wrong, would appreciate your help.
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first solution – let's choose the person that hits the 5th floor such that the elevator will reach this floor. $\binom{5}{1}$ options for that. for the rest, there're $5^4$ options to choose a floor. the probability space is $10^5$ for number of ways to choose floors (out of 10 floors) independently for 5 people. so the final result would be $\frac{\binom{5}{1} 5^4}{10^5}$
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second solution – let's notice there're $5^5$ ways to hit buttons in elevator such that the elevator would get at the most to 5th floor, and $4^5$ ways to hit buttons in elevator such that the elevator would get at the most to 4th floor. so the number of ways the elevator would reach exactly the 5th floor is $5^5-4^5$ and then the result is $\frac{5^5-4^5}{10^5}$
Best Answer
I think its easier to understand the difference if you simplify the example.
Assume you have 4 floors, 2 people, and you want to get the probability that the elevator would get exactly to the 2nd floor and no higher.
second solution would be $$ \frac{2^2 - 1 ^2}{4^2} = \frac{3}{16}, $$because there are 16 possible scenarios but only 3 lead exactly to the 2nd floor (namle 1-2, 2-1 and 2-2).
first solution You get $$ \frac{\binom{2}{1} 2^1}{4^2} = \frac{4}{16} $$ because you count that the first persons hits floor 2 and the second person 1 or 2 (namely 2-1 and 2-2) plus the other case (namely 1-2 and 2-2). But you got the case 2-2 twice and this is why your first solution does not work.