
Probability and Statistics Basic Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
March 24th, 2013, 02:44 PM  #1 
Newbie Joined: Mar 2013 Posts: 7 Thanks: 0  help with probability please
B and R are members of a 15person club that selects a president and his assistant by lottery. What is the probability that B will be a president and R his assistant? (one person cannot hold 2 positions) 
March 24th, 2013, 05:08 PM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024  Re: help with probability please
2/15 * 1/14 (if 2 names drawn)

March 24th, 2013, 05:17 PM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,934 Thanks: 1128 Math Focus: Elementary mathematics and beyond  Re: help with probability please
Why 2/15?

March 24th, 2013, 05:32 PM  #4  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024  Re: help with probability please Quote:
2nd draw (if 1st B or R): 14 left; so 1/14 Thassa way I read the problem...order does not count? No?  
March 24th, 2013, 05:37 PM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,934 Thanks: 1128 Math Focus: Elementary mathematics and beyond  Re: help with probability please
Draw for president: 1/15. Then draw for assistant: 1/14. 1/15 * 1/14 = 1/210. Draw for assistant: 1/15. Then draw for president: 1/14. 1/15 * 1/14 = 1/210. 
March 24th, 2013, 06:11 PM  #6 
Newbie Joined: Mar 2013 Posts: 7 Thanks: 0  Re: help with probability please
Yes, order does not count, so I figured I'd use the formula for permutations: Pn,r= n!/(nr)! vs Cn,r=n!/r!(nr)! (which is for problems where order doesn't matter) I know that probability is counted as follows: P(E)= n(E)/n(S); in this case n(E) is number of elements in a compound event that can be broken down into two: first, we choose B as the president P (given 15, choose 1) or, in other words, P15,1= 15!/14!=15 second, we choose R as assistant out of 14, since B is already a president: P (given 14, choose 1) or P14,1=14 Using multiplication principle we find out all the possible combinations: 14x15=210 Now, to calculate the probability, we need to divide the number of elements in the compound even (210) by the sample space (S) which is the list of all possible combinations... Isn't the sample space P15,2 (given 15, choose 2)??? So I divided 210 by 210 and got 1 which is unrealistic given the conditions of the problem... Where's my mistake? 
March 24th, 2013, 06:16 PM  #7 
Newbie Joined: Mar 2013 Posts: 7 Thanks: 0  Re: help with probability please
***Oops, my bad, it's the other way around. Here order matters because B must be the president and R his assistant, so I used the P formula as opposed to the C formula for where order does not matter. Anyway, where's the mistake? 
March 24th, 2013, 07:51 PM  #8 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024  Re: help with probability please
IF that's the case, then problem should be clear and state: 1st draw = president, 2nd draw = assistant. The way it reads now is: 2 names are drawn; if they're B and R, then B will be president and R the assistant. 
March 24th, 2013, 10:46 PM  #9 
Newbie Joined: Mar 2013 Posts: 7 Thanks: 0  Re: help with probability please
Well, the problem reads exactly as follows: "Brittani and Ramon are members of a 15person ski club. If a president and assistant selected by lottery, what is the probability that Brittani will be president and Ramon will be assistant? ( A person cannot hold more than one office)" 

Tags 
probability 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
probability set theory  probability of x out of n events  hbonstrom  Applied Math  0  November 17th, 2012 07:11 PM 
Joint probability density function, probability  token22  Advanced Statistics  2  April 26th, 2012 03:28 PM 
Probability (probability mass function,pmf)  naspek  Calculus  1  December 15th, 2009 01:18 PM 