probability problem - XP Math - Forums

 XP Math - Forums probability problem

 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 #5 Temperal Guest   Posts: n/a Haha, forgot we're not on AoPS, did you, Siddhant?
 08-01-2007 #6 Sillysidley   Join Date: Oct 2006 Posts: 822 Yeah, same thing happened on my old post... thanks for fixing it __________________ .

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules
 Forum Jump User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home Welcome     XP Math News     Off-Topic Discussion Mathematics     XP Math Games Worksheets     Homework Help     Problems Library     Math Challenges

All times are GMT -4. The time now is 11:40 AM.

 Contact Us - XP Math - Forums - Archive - Privacy Statement - Top