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Math Help - Probability counting excercise

  1. #1
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    Probability counting excercise

    Ten cards are drawn randomly from a deck of 52 cards (13 cards of each of the 4 different suits).

    Each selected card is put into a pile that depends on what suit the card is. (4 total piles)

    a) What is the probability that the largest pile contains 4 cards, the next largest pile has 3 cards, the next largest has two cards and the smallest has 1 card?

    I am not sure how to begin part a. You want to select 4 of one suit, 3 of a different suit, then 2 of a yet again different suit and then a single card of the last suit. If the numbers were tied to specific suits I could see how to do part a, but this more general case is confusing me.

    b) What is the probability that two of the piles have 3 cards, one has 4 cards, and one has no cards?

    So when you draw ten cards you don't get 1 of the 4 suits. However, like part a, since the numbers aren't tied to specific suits, I am confused on how to proceed.

    Any help would be greatly appreciated. Thank you kindly.
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  2. #2
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    a) Do you suppose you can solve the problem if given the suits? Say 4 hearts, 3 clubs, 2 diamonds, and 1 spade?

    If so, then just multiply your answer by 4! to allow for the possible arrrangements of the suits.
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  3. #3
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    so the answer to part a is

    \frac{4!*\begin{pmatrix}\;\;13<br />
&\\\;\;4<br />
\end{pmatrix}*\begin{pmatrix}\;\;13<br />
&\\\;\;3<br />
\end{pmatrix}\begin{pmatrix}\;\;13<br />
&\\\;\;2<br />
\end{pmatrix}\begin{pmatrix}\;\;13<br />
&\\\;\;1<br />
\end{pmatrix}}{\begin{pmatrix}\;\;52<br />
&\\\;\;10<br />
\end{pmatrix}}

    two questions

    why isn't the denominator 52*51*50*49*48*47*46*45*44*43

    also in part b do we multiply by 3! or still by 4! since we still have to consider the pile of zero cards?
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  4. #4
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    Hello, Sheld!

    Here's another approach . . . I'll do part (a).


    Ten cards are drawn randomly from a deck of 52 cards.
    Each selected card is put into a pile that depends on what suit the card is.

    (a) What is the probability that the largest pile contains 4 cards, the next largest
    has 3 cards, the next largest has 2 cards and the smallest has 1 card?

    There are: . \displaystyle{52\choose10} possible 10-card hands.
    . . This is the denominator of the probability.

    It is not _{52}P_{10} because the order of the cards is not considered.


    For the first pile, there are 4 choices for the suit.
    . . Then there are {13\choose4} ways to get 4 cards of that suit.

    For the second pile, there are 3 choices for the suit.
    . . Then there are {13\choose3} ways to get 3 cards of that suit.

    For the third pile, there are 2 choices for the suit.
    . . Then there are {13\choose2} ways to get 2 cards of that suit.

    For the fourth pile, there is 1 choice for the suit.
    . . Then there are {13\choose1} ways to get 1 card of that suit.

    Hence, the numerator is: . \displaystyle 4\!\cdot\!{13\choose4}\!\cdot\! 3\cdot\!{13\choose3}\!\cdot\!2\cdot\!{13\choose2}\  !\cdot\!1\!\cdot\!{13\choose1}

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  5. #5
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    Ah I understand now. The 10 cards are the 10 cards, it doesnt matter what order they are in.

    I also believe I got the answer to part b)

    \frac{4!*\begin{pmatrix}\;\;13<br />
&\\\;\;4<br />
\end{pmatrix}*\begin{pmatrix}\;\;13<br />
&\\\;\;3<br />
\end{pmatrix}\begin{pmatrix}\;\;13<br />
&\\\;\;3<br />
\end{pmatrix}\begin{pmatrix}\;\;13<br />
&\\\;\;0<br />
\end{pmatrix}}{\begin{pmatrix}\;\;52<br />
&\\\;\;10<br />
\end{pmatrix}}

    where
    \begin{pmatrix}\;\;13<br />
&\\\;\;0<br />
\end{pmatrix} = 1

    Thanks a lot!
    Last edited by Sheld; March 8th 2011 at 08:52 PM. Reason: latex formatting
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