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TOPIC: Please Help Solve This Probability Problem!! 


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Joined Apr 2011 Posts: 224 
September 17, 2013 9:18 AM
Can someone please show me what formula needs to be used to solve the problem below? I know the answer is 1/228 but please show me what formula, etc. they used to come up with that particular answer. Thanks!
In a sample bag of M&M's candy, there are 5 brown, 6 yellow, 4 blue, 3 green, and 2 orange. What is the probability of getting 1 brown and 2 orange M&M's if 3 are taken at a time? The answer is c. 1/228 
Joined Jan 2013 Posts: 790 
September 17, 2013 9:52 AM
Suppose the first choice is brown. 5/20. Next choices are orange. 2/19, 1/18. Multiply these together.
Suppose instead that the first choice is orange. 2/20. Then it's either orange and brown, 1/19 times 5/18, OR brown and orange, 5/19 times 1/18. These alternatives are added, and the result multiplied by the probability of the first choice being orange. Total probability: (5/20 * 2/19 * 1/18) + (2/20 * ((1/19 * 5/18) + (5/19 * 1/18))) = 1/684 + (1/10 * 10/342) = 3/684 = 1/228 or 0.4386 % 
Joined Nov 2008 Posts: 9,547 
September 17, 2013 9:58 AM
It made my head hurt but can I still have a cookie?

Joined Feb 2012 Posts: 482 
September 17, 2013 10:02 AM
QUOTE: Suppose the first choice is brown. 5/20. Next choices are orange. 2/19, 1/18. Multiply these together. Suppose instead that the first choice is orange. 2/20. Then it's either orange and brown, 1/19 times 5/18, OR brown and orange, 5/19 times 1/18. These alternatives are added, and the result multiplied by the probability of the first choice being orange. Total probability: (5/20 * 2/19 * 1/18) + (2/20 * ((1/19 * 5/18) + (5/19 * 1/18))) = 1/684 + (1/10 * 10/342) = 3/684 = 1/228 or 0.4386 % Correct. This is just like when I took the LSATS for law school. Except I'd need to figure out what order the colors would be in 
Joined Feb 2012 Posts: 230 
September 17, 2013 10:03 AM
WTF?

Joined Apr 2012 Posts: 1,186 
September 17, 2013 10:11 AM
*blank stare*
Yup, head hurts. 
Joined Apr 2011 Posts: 224 
September 17, 2013 10:19 AM
Thanks so much for your help Christy! :)

Joined Jan 2013 Posts: 790 
September 17, 2013 10:21 AM
QUOTE: Thanks so much for your help Christy! :) You're welcome! 
Joined Mar 2011 Posts: 1,842 
September 17, 2013 10:24 AM
QUOTE: Suppose the first choice is brown. 5/20. Next choices are orange. 2/19, 1/18. Multiply these together. Suppose instead that the first choice is orange. 2/20. Then it's either orange and brown, 1/19 times 5/18, OR brown and orange, 5/19 times 1/18. These alternatives are added, and the result multiplied by the probability of the first choice being orange. Total probability: (5/20 * 2/19 * 1/18) + (2/20 * ((1/19 * 5/18) + (5/19 * 1/18))) = 1/684 + (1/10 * 10/342) = 3/684 = 1/228 or 0.4386 % I think I'm in love... 
Joined Jan 2012 Posts: 1,661 
September 17, 2013 10:27 AM
I did it similar to Christy.
Your possible orders are OOB, OBO, or BOO (three of them). Total possible M&Ms are 20. 3*(5/20*2/19*1/18) = 30/6840 = 1/228 
Joined Jan 2013 Posts: 790 
September 17, 2013 11:16 AM
QUOTE: QUOTE: Suppose the first choice is brown. 5/20. Next choices are orange. 2/19, 1/18. Multiply these together. Suppose instead that the first choice is orange. 2/20. Then it's either orange and brown, 1/19 times 5/18, OR brown and orange, 5/19 times 1/18. These alternatives are added, and the result multiplied by the probability of the first choice being orange. Total probability: (5/20 * 2/19 * 1/18) + (2/20 * ((1/19 * 5/18) + (5/19 * 1/18))) = 1/684 + (1/10 * 10/342) = 3/684 = 1/228 or 0.4386 % I think I'm in love... How YOU doin? 
Joined Sep 2013 Posts: 887 
September 17, 2013 11:24 AM
42.
At least that's what my Infinite Improbability drive tells me. 
Joined Jan 2013 Posts: 790 
September 17, 2013 11:28 AM
QUOTE: 42. At least that's what my Infinite Improbability drive tells me. you look delicious today! 
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