if someone have 4 red bricks and 8 blue bricks ,he want to arrange them randomly in circular shape,What is the probabilitythat two red bricks are notside by side?

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- Aug 16th 2012, 08:04 AMMhmh96the 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*? - Aug 16th 2012, 10:15 AMSorobanRe: the probability of bricks arranged randomly.
Hello, Mhmh96!

Quote:

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 areside by side?*not*

There are $\displaystyle \frac{12!}{4!\,8!} \,=\, 495$ possible arrangements.

(This includes*duplication from rotations*.)

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

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

Select 4 of the 8 spaces and insert the red bricks.

There are: $\displaystyle \frac{8!}{4!\,4!} \,=\,70$ ways.

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