XP Math Games Arcade

07-31-2007

Hello!
I cann't think out this problem.Please help me.ok?
There is a possible stop at every floor(floor 3,floor 2 and floor 1).Assume that any person is likely to get off at any floor and let X be the number of people that get off on the first floor.
(i)What values can X take on?
(ii)Find the probability for each possible value of X.
(iii) What is the expected number of people getting off on the first floor?

**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...

**Scion** · 08-01-2007

I wouldn't be helping much since somedude and i share the same answers. Lets say there are Y people.
1)If they have an equal chance of getting off the answer to this would simply be anywhere from 0 to Y(sidenote: if i said between 0 and Y that would not include 0 and Y).
2)I thought when you kind of answered the question in the information given since they have an equal chance so everything would have a $\frac{1}{Y}$ chance since theres an equal probability.
3)straight-forward $\frac{Y}{3}$ this is because if theres an equal chance we would expect equal people to get off at each floor and since there are Y people the answer was $\frac{Y}{3}$ .

**Sillysidley** · 08-01-2007

I don't agree with your number two. Out of the Y people, each one has a $\frac{1}{3}$ probability of going to floor one and a $\frac{2}{3}$ chance of not going on floor one. The only thing I'm not sure of, is since we choose 1 of the people to be on floor one, I think we might have to multiply by Y since we have Y ways to choose that Y person...

08-01-2007

Haha, forgot we're not on AoPS, did you, Siddhant?

**Sillysidley** · 08-01-2007

Yeah, same thing happened on my old post... thanks for fixing it

07-31-2007	#1
ray2won Guest Posts: n/a	probability problem Hello! I cann't think out this problem.Please help me.ok? There is a possible stop at every floor(floor 3,floor 2 and floor 1).Assume that any person is likely to get off at any floor and let X be the number of people that get off on the first floor. (i)What values can X take on? (ii)Find the probability for each possible value of X. (iii) What is the expected number of people getting off on the first floor?

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...*

08-01-2007	#3
Scion Join Date: Nov 2006 Posts: 437	I wouldn't be helping much since somedude and i share the same answers. Lets say there are Y people. 1)If they have an equal chance of getting off the answer to this would simply be anywhere from 0 to Y(sidenote: if i said between 0 and Y that would not include 0 and Y). 2)I thought when you kind of answered the question in the information given since they have an equal chance so everything would have a $\frac{1}{Y}$ chance since theres an equal probability. 3)straight-forward $\frac{Y}{3}$ this is because if theres an equal chance we would expect equal people to get off at each floor and since there are Y people the answer was $\frac{Y}{3}$ . __________________

08-01-2007	#4
Sillysidley Join Date: Oct 2006 Posts: 822	I don't agree with your number two. Out of the Y people, each one has a $\frac{1}{3}$ probability of going to floor one and a $\frac{2}{3}$ chance of not going on floor one. The only thing I'm not sure of, is since we choose 1 of the people to be on floor one, I think we might have to multiply by Y since we have Y ways to choose that Y person... __________________ . Last edited by Temperal; 08-01-2007 at 05:46 PM.. Reason: LaTeX

08-01-2007	#6
Sillysidley Join Date: Oct 2006 Posts: 822	Yeah, same thing happened on my old post... thanks for fixing it __________________ .