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TOPIC: Please Help Solve This Probability Problem!!

 
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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
  6488453
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 %
  36293756
September 17, 2013 9:58 AM
It made my head hurt but can I still have a cookie? smile
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 explode
September 17, 2013 10:03 AM
WTF?
  17739223
September 17, 2013 10:11 AM
*blank stare*

Yup, head hurts.
  21742957
September 17, 2013 10:19 AM
Thanks so much for your help Christy! :)
  6488453
September 17, 2013 10:21 AM
QUOTE:

Thanks so much for your help Christy! :)


You're welcome!
  36293756
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...
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
  14795220
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? wink
  36293756
September 17, 2013 11:24 AM
42.

At least that's what my Infinite Improbability drive tells me.
September 17, 2013 11:28 AM
QUOTE:

42.

At least that's what my Infinite Improbability drive tells me.


you look delicious today!
  36293756

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