Results 1 to 15 of 15

Math Help - Probability problem

  1. #1
    Junior Member
    Joined
    Jul 2007
    Posts
    43

    Probability problem

    Yet another brain freeze. "A student is munching on a bag of plain M&Ms. With 22 pieces left, 5 of which are red and 3 of which are yellow, she spills 10 pieces.
    a. Find the probability that the bag still contains the 5 red and 3 yellow pieces.
    b. Determine the probability that exactly 1 of each color red and yellow remain in the bag."

    For a, the answer is simply \frac{7}{11} or .636, correct?

    I can't think of how to approach b.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    For part b.

    We must choose 4 of the 14 'other' colors, 4 of the 5 reds and 2 of the 3 yellows out of 10 chosen from 22.

    I would think:

    \frac{C(14,4)C(5,4)C(3,2)}{C(22,10)}=\frac{15}{646  }\approx{0.0232...}

    That seems rather low, doesn't it?. I may be off base for whatever reason.

    Soroban, Plato, etc. may be along to either concur or prove me wrong.
    Last edited by galactus; July 27th 2007 at 06:35 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,797
    Thanks
    1690
    Awards
    1
    Part a asks for the probability that none of the red or yellow M&Mís are spilled.
    Is that \frac{C(14,10)}{C(22,10)}?

    For part b: \frac{C(14,4)C(5,4)C(3,2)}{C(22,10)}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jul 2007
    Posts
    43
    For guys with all the answers, you sure pose a lot of questions
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Apparently, I was correct. I thought it seemed right, but the probability, intuitively, seemed awful low. Oh well, they can be tricky.

    The first one has a very low probability. One would think it'd be higher, huh?.

    There's another way to look at part a.

    \Pi_{k=0}^{9}\left(\frac{14-k}{22-k}\right)

    part b:

    \frac{10!}{4!4!2!}\Pi_{k=0}^{3}\frac{5-k}{22-k}\cdot\Pi_{k=0}^{1}\frac{3-k}{18-k}\cdot\Pi_{k=0}^{3}\frac{14-k}{16-k}=\frac{15}{646}
    Last edited by galactus; July 27th 2007 at 09:17 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jul 2007
    Posts
    43
    Quote Originally Posted by galactus View Post
    For part b.


    \frac{C(14,4)C(5,4)C(3,2)}{C(22,10)}=\frac{15}{646  }\approx{0.0232...}
    Just curious - what is your technique for simplifying this fraction? In particular, how did you know that you could cancel out the C(14,4) from both parts?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,797
    Thanks
    1690
    Awards
    1
    \frac{{C(14,4)C(5,4)C(3,2)}}{{C(22,10)}} = \frac{{\frac{{14!}}{{4!10!}}\frac{{5!}}{{4!1!}}\fr  ac{{3!}}{{2!1!}}}}{{\frac{{22!}}{{10!12!}}}}<br />
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Actually, I ran it through my calculator to find the answer. I didn't do it by hand. But if you want to, use \frac{n!}{(n-r)!r!}

    For C(14,4), you have \frac{14!}{10!4!}=

    \frac{14*13*12*11*10*9*8*7*6*5*4*3*2}{10*9*8*7*6*5  *4*3*2*4*3*2}=

    Do the cancellations and you get:

    \frac{14*13*12*11}{4*3*2*1}=1001

    Is that what you mean?.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Jul 2007
    Posts
    43
    Quote Originally Posted by Plato View Post
    \frac{{C(14,4)C(5,4)C(3,2)}}{{C(22,10)}} = \frac{{\frac{{14!}}{{4!10!}}\frac{{5!}}{{4!1!}}\fr  ac{{3!}}{{2!1!}}}}{{\frac{{22!}}{{10!12!}}}}<br />
    yes, I understand that's the way to do it... I was just wondering if there's a simple way to know that C(14,4) or 1001 cancels out from the fraction...
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,797
    Thanks
    1690
    Awards
    1
    Quote Originally Posted by earachefl View Post
    I was just wondering if there's a simple way to know that C(14,4) or 1001 cancels out from the fraction...
    I think that one just has to to the work.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Jul 2007
    Posts
    43
    Quote Originally Posted by Plato View Post
    I think that one just has to to the work.
    Bah, humbug!!!
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Did you take note of the other way I showed?. Do you follow it?. Just another way.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Junior Member
    Joined
    Jul 2007
    Posts
    43
    Quote Originally Posted by galactus View Post
    Did you take note of the other way I showed?. Do you follow it?. Just another way.
    well, I saw it, but I don't understand it
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Capital Pi stands for product, just like capital sigma stands for sum.

    the \frac{10!}{4!4!2!}=3150 is the number of ways to arrange the 10 m&m's. Because there are 10 total with 4 red, 2 yellow and 4 others.

    The probability of 4 out of 5 reds is \Pi_{k=0}^{3}\frac{5-k}{18-k}=<br />
(5/22)(4/21)(3/20)(2/19)=\frac{1}{1463}

    The probability of choosing the 2 yellows is \Pi_{k=0}^{1}\frac{3-k}{18-k}=(3/18)(2/17)=\frac{1}{51}

    The probability of choosing the 4 others is: \Pi_{k=0}^{3}\frac{14-k}{16-k}=(14/16)(13/15)(12/14)(11/13)=\frac{11}{20}

    3150(1/1463)(1/51)(11/20)=15/646
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Junior Member
    Joined
    Jul 2007
    Posts
    43
    well, I'm gonna chew on that one for a while....
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: January 21st 2011, 11:47 AM
  2. Probability Problem
    Posted in the Statistics Forum
    Replies: 3
    Last Post: January 7th 2010, 10:49 PM
  3. Replies: 0
    Last Post: October 8th 2009, 08:45 AM
  4. Probability problem 4
    Posted in the Statistics Forum
    Replies: 1
    Last Post: March 9th 2009, 11:38 PM
  5. A Probability Problem
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: December 27th 2008, 05:17 PM

Search Tags


/mathhelpforum @mathhelpforum