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I want to check that my solution is correct for multinomial probability.

  1. Sep 6, 2011 #1
    1. The problem statement, all variables and given/known data
    A box of candy hearts contains 52 hearts, of which 19 are white, 10 are tan, 7 are pink, 3 are purple, 5 are yellow, 2 are orange, and 6 are green. If you select 9 pieces of candy randomly, without replacement(I don't know why it is in comma's since it is necessary for the probability), give the probability that:

    a) Three of the hearts are white

    b) Three are white, 2 are tan, 1 is pink, 1 is yellow, and 2 are green.

    2. Relevant equations
    I used a multinomial probability for part (b) and the simpler binomial probability for (a). So that (a) looks like:

    (n C r)(p1*p2*p3)(q1*q2*q3*q4*q5*q6) where n=9 for the 9 "trials" . r=3 for the 3 successes. p = successful probability. q = failure probability.

    Now, this is without replacement so the probability changes from p1 to p2 to p3 and, of course q1 to q2 ...

    Now b, I'm not good with the notation using LaTex. But it'd be similar to:

    (n!)/(nw!*nt!*np!*ny!*ng!)*(w1*w2*w3)(t1*t2)(p1)(y1)(g1*g2) where the nw,...,ng represent the number of w,...,number of green. The w1,...g2 represent the respective probabilities (without replacement) of each color.

    3. The attempt at a solution
    The solution went as follows:

    a) (9 C 3)(19/52)(18/51)(17/50)(33/49)(32/48)(31/47)(30/46)(29/45)(28/44) approximately = .292

    b) (9!)/(19!2!1!1!2!)(19/52)(18/51)(17/50)(10/49)(9/48)(7/47)(5/46)(6/45)(5/44) approximately = .006

    To me, these make sense so I'm fairly confident in the solution method. I hope that the approximation is close to the right answer as well. If it is wrong, please let me know!
  2. jcsd
  3. Sep 6, 2011 #2


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    Those answers are correct.

    You can solve the problems a bit more simply by counting the ways you can draw the various combinations. For example, for the first problem, you have 19C3 ways to draw the 3 white hearts, 33C6 ways to draw 6 non-white hearts, and 52C9 ways to draw any 9 hearts, so the probability is (19C3)(33C6)/(52C9). This kind of distribution is called the hypergeometric distribution.
  4. Sep 6, 2011 #3
    Oh wow! Haha, I knew that there was similarity to the method commonly used for poker hands. Where you take 52C5 as your total hand combinations possible and use that to find the probability of any combination of a hand. But we didn't discuss that in class and my book didn't mention even anything more than the multinomial probability. Now I see why! Thank you!

    I have posted another question on the board, and I'd appreciate your help! I'm not looking necessarily for an answer, but I don't know how to think of this problem otherwise.

    Thanks again!
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