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February 23rd, 2012, 05:00 AM  #1 
Senior Member Joined: Jan 2010 Posts: 205 Thanks: 0  Probability of selecting things in order Part 2
Say there are 15 red balls, 10 green balls, and 3 yellow balls within a container. What is the probability that I will select 1 red ball first, then 1 green ball, then 1 more green ball, then 1 red ball, and then 1 yellow ball, in that order? My best guess would be (15 / 2 * (10 / 27) * (9 / 26) * (14 / 25) * (3 / 24) = 1/208 = about 0.48% What would be the permutation way to calculate this? Is 30Permutation5 = 17,100,720 going to be 17,100,720 different ways of picking 5 balls in certain order from the total? Which includes choosing all the same color, but isn't this including choosing 5 yellow balls all in a row, even though there are only 3 of them? The balls are not put back after taking them out. 
February 23rd, 2012, 05:13 AM  #2  
Senior Member Joined: Jul 2011 Posts: 227 Thanks: 0  Re: Probability of selecting things in order Part 2 Quote:
 
February 23rd, 2012, 07:23 AM  #3 
Senior Member Joined: Jan 2010 Posts: 205 Thanks: 0  Re: Probability of selecting things in order Part 2
No, you don't put it back

February 23rd, 2012, 08:44 AM  #4 
Senior Member Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0  Re: Probability of selecting things in order Part 2
should be (15 / 28 * (10 / 27) * (9 / 26) * (14 / 25) * (3 / 24) 
February 23rd, 2012, 08:49 AM  #5 
Senior Member Joined: Jan 2010 Posts: 205 Thanks: 0  Re: Probability of selecting things in order Part 2
Oh yea, sorry, I fixed the calculation. Do you know how to set up the permutation/combination formula for this solution? 
February 23rd, 2012, 09:23 AM  #6  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024  Re: Probability of selecting things in order Part 2 Quote:
 

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