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Thread: Probability Problem please help!

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    Probability Problem please help!

    Please help! I'm not exactly sure where to start. I guess at first by figuring out all of the possible options and then is that it or what? But here's the problem: 6 red beads, 4 white beads, and 1 blue bead are placed in a line in random order. What is the probability that no two neighboring beads are the same color?
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    Quote Originally Posted by sweeetcaroline View Post
    But here's the problem: 6 red beads, 4 white beads, and 1 blue bead are placed in a line in random order. What is the probability that no two neighboring beads are the same color?
    Well your sample space is $\displaystyle S= 3!\times 6!\times4!\times1! $ Now you need to find the event space. If we have $\displaystyle 11$ positions, red can only be in one of the following ways:

    _ R _ R _ R _ R _ R _R

    R _ R _ R _ R _ R _R_

    So we only have $\displaystyle 2!$ choices for red. Now no matter where we place white and blue they are not next to each other, so we have $\displaystyle 5! $ ways of placing the remaining $\displaystyle 5$ beads.

    That is $\displaystyle E= 2!\times5!$

    $\displaystyle P= \frac{E}{S} = \frac{2!\times5!}{3!\times 6!\times4!\times1!}$
    Last edited by Anonymous1; Mar 31st 2010 at 06:11 PM.
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    Thanks so much but why in the sample space is 3! included? Wouldn't it just be 6! x 4! x 1!?
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    _ _ _ _ _ _
    We arranged the six red

    _ _ _ _
    Then the four white

    _
    Then the one blue.

    Now think of each one of these color blocks as its own entity. Then we have 3! ways of arranging the 3 entities.

    WAIT!

    S=11!
    Because we have 11! ways of arranging the 11 items.

    Sorry about that.
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    Hello, sweeetcaroline!

    Edit: Plato is absolutely right . . . *blush*


    6 red beads, 4 white beads, and 1 blue bead are placed in a line in random order.
    What is the probability that no two neighboring beads are the same color?
    There are: .$\displaystyle {11\choose6,4,1} \:=\:2310$ possible orders.



    Place the 6 Red beads in a row, leaving a space between them.

    . . $\displaystyle \begin{array}{ccccccccccccc} R & \_ & R & \_ & R & \_ & R & \_& R & \_ & R\end{array}$


    Place the Blue bead is any of the 5 spaces: .$\displaystyle 5$ choices.


    Drop the 4 White beads in the remaining 4 spaces: .$\displaystyle 1$ way.


    Hence, there are: .$\displaystyle 5\cdot1 \,=\,5$ ways.


    Therefore, the probability is: . $\displaystyle \frac{5}{2310} \;\;=\;\;\frac{1}{462}$

    Last edited by Soroban; Apr 1st 2010 at 07:52 AM.
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    Quote Originally Posted by sweeetcaroline View Post
    the problem: 6 red beads, 4 white beads, and 1 blue bead are placed in a line in random order. What is the probability that no two neighboring beads are the same color?
    Can someone give me a line of these beads in which no two neighboring beads are the same color other than the five listed below?
    $\displaystyle RWRWRWRWRBR$
    $\displaystyle RWRWRWRBRWR$
    $\displaystyle RWRWRWBWRWR$
    $\displaystyle RWRBRWRWRWR$
    $\displaystyle RBRWRWRWRWR$
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