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October 18th, 2013, 11:02 PM  #1 
Newbie Joined: Oct 2013 Posts: 10 Thanks: 0  Basic probability
An urn contains 7 red marbles labeled {1,2,3,4,5,6,7} and 5 green marbles labeled {1,2,3,4,5}. Four marbles are pulled out at once (i.e. with no particular order). What is the probability that (a) all four marbles are red? (b) more of the marbles are green than red? (c) both red and green marbles are present? (d) two of the marbles chosen are both labeled "5"? 
October 19th, 2013, 04:09 PM  #2 
Newbie Joined: Oct 2013 Posts: 10 Thanks: 0  Re: Basic probability
bump

October 19th, 2013, 05:22 PM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: Basic probability Hello, xsgx! Quote:
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October 19th, 2013, 06:03 PM  #4 
Newbie Joined: Oct 2013 Posts: 10 Thanks: 0  Re: Basic probability
Wow, thank you! Just what I needed, a step by step explanation.

October 21st, 2013, 05:18 AM  #5 
Newbie Joined: Oct 2013 Posts: 9 Thanks: 0  Re: Basic probability
Very neat solution. My mind did not think as laterally as you did. By thinking "aloud", I could do (a), the first part of (b) and (c) with some luck. For (a), I did 7/12 x 6/11 x 5/10 x 4/9 = =7/99 For straightforward 4 greens, I got 1/99; I was then faced with doing all the combinations of greens and red. Well done. 
October 21st, 2013, 09:31 PM  #6 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 635 Thanks: 96 Math Focus: Electrical Engineering Applications  Re: Basic probability
[color=#0000BF]nubian123[/color], Using your method, the probabilities of the various combinations of greens and reds are not too hard to calculate: The factorial expressions multiplied at the end are the number of permutations in the order that the marbles may be picked. For example, with 1 green and 3 red; GRRR, RGRR, RRGR, RRRG gives 4. For 2 green, 2 red; RRGG, RGRG, RGGR, GRRG, GRGR, GGRR gives 6. Notice that all probabilities add to 1 as they should: d) is worked as: 

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