# Thread: Probability Problem please help!

1. ## 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?

2. Originally Posted by sweeetcaroline
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 $S= 3!\times 6!\times4!\times1!$ Now you need to find the event space. If we have $11$ 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 $2!$ choices for red. Now no matter where we place white and blue they are not next to each other, so we have $5!$ ways of placing the remaining $5$ beads.

That is $E= 2!\times5!$

$P= \frac{E}{S} = \frac{2!\times5!}{3!\times 6!\times4!\times1!}$

3. Thanks so much but why in the sample space is 3! included? Wouldn't it just be 6! x 4! x 1!?

4. _ _ _ _ _ _
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.

5. 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: . ${11\choose6,4,1} \:=\:2310$ possible orders.

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

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

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

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

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

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

6. Originally Posted by sweeetcaroline
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?
$RWRWRWRWRBR$
$RWRWRWRBRWR$
$RWRWRWBWRWR$
$RWRBRWRWRWR$
$RBRWRWRWRWR$