Results 1 to 6 of 6

Math Help - Probability

  1. #1
    Newbie
    Joined
    Jun 2009
    Posts
    8

    Probability

    We have 8 bottles of Zinfandel. 10 of Merlot, and 12 of Cabernet.

    If he wants to serve 3 bottles of zinfandel and serving order is important, how many ways are there to do this?
    P=8!/(8-3)! = 336

    If 6 bottles of wine are to be randomly selected for the 30 for serving, how many ways are there to do this?
    P=30!/6!(30-6)! = 593,775

    If 6 bottles are randomly selected, how many ways are there to obtain two bottles of each variety?
    P = {{\binom{8}{2}} {\binom{12}{2}} {\binom{10}{2}}} = 83,160<br />

    If 6 bottles are randomly selected, what is the probability that this results in two bottles of each variety being chosen?
    P = {{\binom{8}{2}} {\binom{12}{2}} {\binom{10}{2}}}/{\binom{30}{6}} = .1400<br />

    If 6 bottles are randomly selected, what is the probability that all of them are the same variety.
    Not even sure how to start this one. Any help?

    Are these right?
    Last edited by hansel13; June 28th 2009 at 12:44 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,966
    Thanks
    1785
    Awards
    1
    The last one:
    \displaystyle \frac{\binom{8}{6}+\binom{10}{6}+\binom{12}{16}}{\  binom{30}{6}}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2009
    Posts
    8
    Quote Originally Posted by Plato View Post
    The last one:
    \displaystyle \frac{\binom{8}{6}+\binom{10}{6}+\binom{12}{16}}{\  binom{30}{6}}
    How did you get 6, 6, and 16?

    Also, I have another general question. When should I use the following formulas:
    P = n!/(n-k)!
    and
    P = n!/(n-k)!*k!

    Neither my book/teacher really explain the difference between these two. Is the first just the number of combinations, and the second a percentage of possibilities?

    thanks, btw is #4 really correct? I think I get a different number than the answer, much appreciated!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,966
    Thanks
    1785
    Awards
    1
    Quote Originally Posted by Plato View Post
    The last one:
    \displaystyle \frac{\binom{8}{6}+\binom{10}{6}+\binom{12}{16}}{\  binom{30}{6}}
    It should have been \displaystyle \frac{\binom{8}{6}+\binom{10}{6}+\binom{12}{6}}{\b  inom{30}{6}}
    It was a typo.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jun 2009
    Posts
    8
    Quote Originally Posted by Plato View Post
    It should have been \displaystyle \frac{\binom{8}{6}+\binom{10}{6}+\binom{12}{6}}{\b  inom{30}{6}}
    It was a typo.
    thank you, also does 4 seem correct?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,966
    Thanks
    1785
    Awards
    1
    Quote Originally Posted by hansel13 View Post
    thank you, also does 4 seem correct?
    YES
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: January 21st 2011, 12:47 PM
  2. Replies: 3
    Last Post: May 29th 2010, 08:29 AM
  3. fundamental in probability & convergence with probability 1
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: February 23rd 2010, 10:58 AM
  4. Replies: 1
    Last Post: February 18th 2010, 02:54 AM
  5. Replies: 3
    Last Post: December 15th 2009, 07:30 AM

Search Tags


/mathhelpforum @mathhelpforum