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Math Help - the probability of bricks arranged randomly.

  1. #1
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    the probability of bricks arranged randomly.

    if someone have 4 red bricks and 8 blue bricks ,he want to arrange them randomly in circular shape, What is the probability that two red bricks are not side by side?
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  2. #2
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    Re: the probability of bricks arranged randomly.

    Hello, Mhmh96!

    If someone have 4 red bricks and 8 blue bricks and arranges them randomly in a circle,
    what is the probability that two red bricks are not side by side?

    There are \frac{12!}{4!\,8!} \,=\, 495 possible arrangements.
    (This includes duplication from rotations.)


    Place the 8 blue bricks in a circle, leaving a space between them.

    . . \begin{array}{cccccccc}&& B & \circ & B \\ & \circ &&&& \circ \\ B &&&&&& B \\ \circ &&&&&& \circ \\ B &&&&&& B \\ & \circ &&&& \circ \\ && B & \circ & B \end{array}


    Select 4 of the 8 spaces and insert the red bricks.
    There are: \frac{8!}{4!\,4!} \,=\,70 ways.


    Therefore: . P(\text{Reds non-adjacent}) \:=\:\frac{70}{495} \:=\:\frac{14}{99}

    Thanks from Mhmh96
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