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Math Help - Combinatorial Probability

  1. #1
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    Combinatorial Probability

    Hello, I'm a first time user here. Hopefully someone can help me out...

    At a picnic, there was a bowl of chocolate candy that had 10 pieces each of Milky Way, Almond Joy, Butterfinger, Nestle Crunch, Snickers, and Kit Kat. Jen grabbed six pieces at random from this bowl of 60 chocolate candies.

    a. What is the probability that she got one of each variety?

    b. What is the probability that Jen grabs exactly five varieties?

    I understand question a. I realize that there are 50,063,860 possible outccomes in which this can occur and that there is 1,000,000 ways for Jen to do this. (1,000,000/50,063,860) = probability.

    Having a hard time setting up question b. and ultimately solving it.
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  2. #2
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    Re: Combinatorial Probability

    Quote Originally Posted by kangta27 View Post
    At a picnic, there was a bowl of chocolate candy that had 10 pieces each of Milky Way, Almond Joy, Butterfinger, Nestle Crunch, Snickers, and Kit Kat. Jen grabbed six pieces at random from this bowl of 60 chocolate candies.
    a. What is the probability that she got one of each variety?
    b. What is the probability that Jen grabs exactly five varieties?
    I understand question a. I realize that there are 50,063,860 possible outccomes
    Where in the world did you get that?
    \binom{6+6-1}{6}=462 possible multi-selects of six.

    Of those there is only 1 way to get one of each.

    Of the 462 there is only 5 ways to get only five varieties.
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  3. #3
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    Re: Combinatorial Probability

    well for question a., I used (60C6) which gave me 50,063,860. And then to get one of each: (10C1)^6 which gives me 1,000,000. The answer to question a. is 1,000,000/50,063,860 = 0.01997
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  4. #4
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    Re: Combinatorial Probability

    Quote Originally Posted by kangta27 View Post
    well for question a., I used (60C6) which gave me 50,063,860. And then to get one of each: (10C1)^6 which gives me 1,000,000. The answer to question a. is 1,000,000/50,063,860 = 0.01997
    But what you fail to understand is that this question is about varieties. There are only six varieties.

    If you had to make up a packets of six candies there would be only 462 possible different packets. Of those packets only one would contain all six kinds of candy. And only six contain exactly five kinds.

    You see that all ten of each of variety are considered identical.
    All ten Milky Ways are considered identical.
    But the variety Milky Way is different from the variety Almond Joy.
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    Re: Combinatorial Probability

    Plato, http://www.mathhelpforum.com/math-he...ty-159191.html , this is where I got help for question a. The back of my book also states that 0.01997 is the answer the question a.

    I posted this question so I could get help with question b., if you could help I'd greatly appreciate it! Thanks!
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    Re: Combinatorial Probability

    Quote Originally Posted by kangta27 View Post
    Plato, http://www.mathhelpforum.com/math-he...ty-159191.html , this is where I got help for question a. The back of my book also states that 0.01997 is the answer the question a.
    I posted this question so I could get help with question b., if you could help I'd greatly appreciate it! Thanks!
    That solution is incorrect. This problem comes out of this project.
    If I remember correctly the problem is discussed in Laurie Snell's probability. If you go to that website and then to Laurie's page, I think you can download a PDF of the book.
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  7. #7
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    Re: Combinatorial Probability

    Hello, kangta27!

    I agree with your answer to (a).


    At a picnic, there was a bowl of chocolate candy that had 10 pieces each of:
    Milky Way, Almond Joy, Butterfinger, Nestle Crunch, Snickers, and Kit Kat.
    Jen grabbed six pieces at random from this bowl of 60 chocolate candies.

    (a) What is the probability that she got one of each variety?

    (b) What is the probability that Jen grabs exactly five varieties?

    There are {6\choose5} = {\color{blue}6} ways to choose the five varieties.

    We want one each of four varieties and two of the fifth variety.
    There are 5 choices for the variety from which two pieces are drawn.

    There are: . {10\choose1}^4{10\choose2} ways to select the 6 pieces.


    \displaystyle P(\text{5 varieties}) \:=\:\dfrac{6\cdot5\cdot{10\choose1}^4{10\choose2}  }{{60\choose6}} \;=\; \dfrac{13,\!500,\!000}{50,\!063,\!860} \:\approx\:0.27

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  8. #8
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    Re: Combinatorial Probability

    Soroban, the answer in the back of the book states that answer is 0.05393. I understand that we need two of the 5th variety of chocolate but am having a hard time seeing the set-up. Thanks!
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