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May 14th, 2013, 12:12 AM  #1 
Newbie Joined: May 2013 Posts: 2 Thanks: 0  A probability quiz
First time in this forum so forgive me if this kind of problem has already been discussed or posted! The following question does not seem to be tremendously complex but in discussion with friends/colleagues there seems to be 2 lines of thought. I am sure there is only 1 valid answer though! 40 people (one of them being you) are to be distributed in 10 groups of 4 people each. The only restriction is that 5 of those people (you not included in those 5) need to be in separate groups. What is the probability for you being in a group with any of those 5? Line of thought 1: lets first calculate the probability of each of those 5 not to be in the group you are in: 9/10 for the 1st, 8/9 for the second, 7/8 for the 3rd, 6/7 for the 4th and 5/6 for the 5th. Then you calculate 1  the above probability for the chance to actually have one of those 5 in your group, so 1(9/10*8/9*7/8*6/7*5/6)= 0.5. Line of thought 2: There are 35 people to be distributed in the 35 spots available other than the 5 occupied by the dreaded 5. The 5 groups with one of the 5 dreaded individuals in them will have 3 x 5 = 15 spots available. Therefore the chances for any of the 35 to be in one of them would be 15/35 = 0.42. Which one is right? 
May 14th, 2013, 05:13 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond  Re: A probability quiz
There are 10 groups and 5 of them contain one special member each. Since the groups are of equal size, the probability that you will end up in any given group is the same, hence the probability you will be in a group with a special member is 1/2.

May 14th, 2013, 05:45 AM  #3  
Newbie Joined: May 2013 Posts: 2 Thanks: 0  Re: A probability quiz Quote:
 

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