Please Help Solve This Probability Problem!!

Metamorphasis555
Metamorphasis555 Posts: 224
in Chit-Chat
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

Replies

  • WhatAnAss
    WhatAnAss Posts: 1,598 Member
    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 %
  • Carl01
    Carl01 Posts: 9,307 Member
    It made my head hurt but can I still have a cookie? :smile:
  • jess7386
    jess7386 Posts: 477 Member
    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:
  • mikeschratz
    mikeschratz Posts: 253 Member
    WTF?
  • AmyP619
    AmyP619 Posts: 1,137 Member
    *blank stare*

    Yup, head hurts.
  • Metamorphasis555
    Metamorphasis555 Posts: 224
    Thanks so much for your help Christy! :)
  • WhatAnAss
    WhatAnAss Posts: 1,598 Member
    Thanks so much for your help Christy! :)

    You're welcome!
  • baldzach
    baldzach Posts: 1,841 Member
    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...
  • dbrightwell1270
    dbrightwell1270 Posts: 1,732 Member
    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
  • WhatAnAss
    WhatAnAss Posts: 1,598 Member
    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:
  • RuleFive
    RuleFive Posts: 880 Member
    42.

    At least that's what my Infinite Improbability drive tells me.
  • WhatAnAss
    WhatAnAss Posts: 1,598 Member
    42.

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

    you look delicious today!

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