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There is a 10 floors building, 10 people get in the elevator in the ground floor and get off at each floor independently. What is the probability that the elevator stops at floor 5?

My answer is that because there are 10 floors so the probability that the elevator stops at floor 5 (same as floor 1,2,3,4..) is 1 - (9/10)

Is my answer correct here? I think it's too simple for this question.

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    $\begingroup$ Your answer is too simple, and shouldn't survive a simple sanity check. $\endgroup$ – Zackkenyon Nov 11 '13 at 21:15
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The way I view this, there are $10$ floors, each person getting into an elevator has a probability of getting off on the 5th floor of $1/10$ and not getting off there of $9/10$. The probability of all $10%$ people not getting off on the 5th floor is $(9/10)^{10}$; therefore, the probability that the elevator stops on the fifth floor at all is the probability that at least one person gets off on the 5th floor, or

$$1-\left ( \frac{9}{10}\right)^{10} \approx 0.6513$$

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  • $\begingroup$ thank you for your clarification $\endgroup$ – Squall Leonahart Nov 11 '13 at 21:19
  • $\begingroup$ @SquallLeonahart: you're welcome. If the answer if useful to you, please consider accepting the answer by clicking on the checkmark to the left. $\endgroup$ – Ron Gordon Nov 11 '13 at 21:39
  • $\begingroup$ It's a $10$ floor building. Every one gets on at the ground floor, so that leaves $9$ floors at which the passengers get off randomly. Or is this the European "ground floor, first floor, second floor.." convention> $\endgroup$ – DJohnM Nov 11 '13 at 22:34

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