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**Sillysidley** · 07-31-2007

Doesn't it depend on how many people there are? Well, I guess I'll let that be Y. Also, I assume you start at the floor below the first.
(i) The possible values range from 0 to Y
(ii) For 0- $(\frac{2}{3})^{X}$ Basically each person has a $\frac{2}{3}$ chance of not going off at floor 1, and there are X people.
Similarly, for 1 (I'm kind of guessing on this) it's $\frac{1}{3}*(\frac{2}{3})^{X-1}$ . ( Or should that be multiplied by Y?)That's a hint, I don't want to do it all, so the rest is yours
(iii)You expect $\frac{1}{3}$ of the poeple to get off at each floor, so the answer is $\frac{Y}{3}$

Hopefully I got some of these right...

07-31-2007	#2
Sillysidley Join Date: Oct 2006 Posts: 822	Doesn't it depend on how many people there are? Well, I guess I'll let that be Y. Also, I assume you start at the floor below the first. (i) The possible values range from 0 to Y (ii) For 0- $(\frac{2}{3})^{X}$ Basically each person has a $\frac{2}{3}$ chance of not going off at floor 1, and there are X people. Similarly, for 1 (I'm kind of guessing on this) it's $\frac{1}{3}(\frac{2}{3})^{X-1}$ . ( Or should that be multiplied by Y?)That's a hint, I don't want to do it all, so the rest is yours (iii)You expect $\frac{1}{3}$ of the poeple to get off at each floor, so the answer is $\frac{Y}{3}$ Hopefully I got some of these right... __________________ . Last edited by Sillysidley; 07-31-2007 at 06:10 PM.. Reason: Thought it was AoPS...*