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Thread: Combination or Permutation

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    Combination or Permutation




    Please see the attached file. I obtained the answer by writing down the outcome sets. I wondered if they were correct. May I ask, how do we compute the said joint probabilities in the problem using combination or permutation? If my answer is wrong, then how do we solve it?

    Thanks.
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    Re: Combination or Permutation

    How did you arrive the probability numbers? You can use Hyper geometric distribution here.
    In the long run, we're all dead.

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    Re: Combination or Permutation

    What is asked is simply the joint. I don't think I have to use the Hyper geometric distribution here. I wrote down the possible outcomes (ie. xxy; xyy; etc..)

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    Re: Combination or Permutation

    Take one possibility say Y= 3 & X=0

    P[Y= 3 & X=0] =P[all three balls are red] =7C3/9C3 which is not same as 1/3
    In the long run, we're all dead.

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    Re: Combination or Permutation

    Just listen to vinux, it is directly related to the hypergeometric distribution.

    Technically speaking you are looking for the multi-hypergeometric distribution; but since you only got 2 categories, therefore it just reduced to the ordinary hypegeometric distribution, just like "Binomial distribution is a special case of the multinomial distribution"

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    Re: Combination or Permutation

    I see. I thought it is just a simple combination. Bdw, when do we use the combination or permutation? Please provide me an answer in your language. I tell you, I've been reading alot of books and browse a multitude of sites already but I'm simply given examples aside from definitions. And the examples are not even sufficient. Thanks.

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    Re: Combination or Permutation

    @vinux, I did try that but when I summed up everything, they did not equal to 1. I will try it again using hypergeometric distribution this time. Thanks.

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    Re: Combination or Permutation

    Hi I did try the hypergeometric distribution. This is what I've got and I've few questions though. Hope you will still entertain.

    The probability that I get 2 white balls and 1 red ball is 1/144. That is,P(X=2)*P(Y=1) where P(X=2)= (3C2)*(6C0)/9C2=1/12; P(Y=1)= (3C1)*(6C6)/9C7=1/12

    Is this correct?

    Also,

    The P(Y=0)= MATH ERROR (same for any Y>3), does this mean I have to put ZERO instead?
    Last edited by dEconomist; 12-30-2011 at 12:24 AM.

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    Re: Combination or Permutation

    Actually, since X + Y \equiv 3, so it is not a joint distribution.

    Of coruse you may express as the following:

    \Pr\{X = x, Y = y\} = \frac {\displaystyle \binom {2} {x} \binom {7} {y}} {\displaystyle \binom {9} {3}} 
= \frac {\displaystyle \binom {2} {x} \binom {7} {3-x}} {\displaystyle \binom {9} {3}} = \Pr\{X = x\}

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    Re: Combination or Permutation

    I don't understand your formula. How do I solve that in calculator? Is that with a C symbol?

    Why not joint? X and Y are both independent so when I get the probability of X AND Y occurring, I just have to multiply their corresponding probability.

    Say, the probability that 2 white balls will occur [P(X=2)] is 1/12 using the hypergeometric; the probability that one red ball will occur [P(Y=1)] is 1/12. So their joint probability will be 1/12*1/12=1/144

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    Re: Combination or Permutation

    X and Y are most definitely not independent. If I tell you X you should very well be able to tell me exactly what Y is.

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    Re: Combination or Permutation

    So what do you suggest? The book explicitly asks for their joint probabilities. How then will I get them? It is easy to provide P(X=2), and P(Y=1) through hypergeometric probability distribution but how do I get their joint probabilities?

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    Re: Combination or Permutation

    Now observe that (X, Y) lies on one straight line - so once either X or Y is given, the other variable is immediately known as a constant. So in this case, it is a degenerated distribution which strictly speaking you cannot say they have a joint distribution.

    Just like you letting X, Y being the number of successes and failures in n i.i.d. Bernoulli trials. Then each of them has a Binomial distribution, but they do not have a joint distribution because again X + Y \equiv n.

    So you need to notice that the multi-hypergeometric distribution/multinomial distribution always has 1 less degree of freedom then the number of random variables. Geometrically, as mentioned above, when there is only 2 random variables, they are perfectly collinear, so it has only 1 dimension and we just treat it as the univariate distribution.

    Back to your question, actually I have already given the formula to you in above, you just need to evaluate the p.m.f. at different values of x is ok.

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    Re: Combination or Permutation

    I will post my answers tomorrow, please check if it's according to your formula. Then if we cannot compute the joint probability, how come the author asked to do it? What do you mean by X + Y \equiv n? Isn't n in hypergeometric is equal to the number of white balls for X and number of red balls for Y? Therefore n(x)=2 and n(y)=7.

    Please enlighten me.

    Thanks.

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    Re: Combination or Permutation


    That is just an arbitrary example - if you are not familiar with multinomial distribution, just forget about it.

    Anyway in your problem you should know that the sum of the number of red balls drawn and the number of white balls drawn is always equal to 3.

    So I say while you are required to find "joint probability", it is actually a little bit misleading because it is a degenerate one - it is not a bivariate distribution, but just like the ordinary univariate distribution. Of course, as I posted before, you can calculate the "joint probability" by the formula I provided, but at the same time it is just equal to the marginal pmf of X or Y alone.

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