Probability question

[yobacul]

Four red balls and six blue balls are placed in a bag. A person takes out the balls one by one at random and places them on a table in the order in which they are extracted. What is the probability that the first two red balls in the row are next to each other?

This is what I am thinking:

(9!*2)/10!

Am I on the right track or do I have to divide by number of ways of arranging the identical objects within themselves. I am not sure.

Thanks in advance for any help.

 

[blamocur]

This is what I am thinking:

(9!*2)/10!

Why? Can you elaborate on how you arrived at this expression?
Thanks.

 

[pka]

Four red balls and six blue balls are placed in a bag. A person takes out the balls one by one at random and places them on a table in the order in which they are extracted. What is the probability that the first two red balls in the row are next to each other
(9!*2)/10!
Am I on the right track or do I have to divide by number of ways of arranging the identical objects within themselves. I am not sure.

If we have $m\text{ red balls}~ \&~n\text{ blue balls}$
there are $\dfrac{(m+n)!}{m!\cdot n!}$ ways to arrange them in a row.
Now we must figure out how many of those have the first two reds next to each other.

 

[yobacul]

Thanks a lot for your help/guidance.

In the denominator I will be arranging all balls in 10! ways but since 4 and 6 are identical, I will divide by 6! and 4! so that in the denominator it would be (10)!/(6!4!) ways.

Now, in the numerator I want 2 red balls to be next to each other. I have 4 red balls in total. So can I group the first two into 1 and the last two into 1. So then I would have to arrange the rest (including the blue ones) in 8! ways. Is my reasoning right? That would give you 192 arrangements.

Thanks a lot

 

[yobacul]

Why? Can you elaborate on how you arrived at this expression?
Thanks.

No, the original working is completely wrong. I have since tried another way but still not sure if it is right or not.

 

[NotJeffM]

No, the original working is completely wrong. I have since tried another way but still not sure if it is right or not.

Please show us that way so we can see if you have done it right on your own or where you slid off the rails.

 

[pka]

Four red balls and six blue balls are placed in a bag. A person takes out the balls one by one at random and places them on a table in the order in which they are extracted. What is the probability that the first two red balls in the row are next to each other?

The question: What is the probability that the first two red balls in the row are next to each other?
In reply #3 I gave you the base number which works out to be
$210$ total ways to arrange ${\color{red}4}\text{ red balls}~\&~{\color{blue}6}\text{ blue balls}$ in a row.
We now are faced with the problem of counting the number of ways that the first to occurrence of two red balls they together.
I find this to be tricky. I suggest that we think of "gluing" two red balls together. Call that twosome $T$.
We are counting the number of ways to arrange $bbbbbbTrr$ so that the $T$ comes before any of the other red balls.
$$

 

[blamocur]

The answer is simple, but the derivation is somewhat hairy: for each $0\leq n \leq 6$ compute the probability $p_n$ of picking exactly $n$ blue balls followed by 2 red balls in a row. The answer then is $P=\sum_{n=0}^6 p_n$.

 

[pka]

The answer is simple, but the derivation is somewhat hairy: for each $0\leq n \leq 6$ compute the probability $p_n$ of picking exactly $n$ blue balls followed by 2 red balls in a row. The answer then is $P=\sum_{n=0}^6 p_n$.

I agree that this is a relatively simple problem to grasp but hard to model.
I see this as basically a counting problem.
There $\dbinom{10}{4}=210$ ways to arrange the row $BBBBBBRRRR$.
Of that number we need to count the number of rows in which the first red ball is followed immediately by a second red ball.
The first red ball can occur in the $k^{th}$ position where $k=1,2,\cdots 7$ That first red can be preceded by $0\text{ to }6$ blue balls.
If the first red is preceded by $j=0,1,\cdots,6$ blue balls then there are $6-j+2$ balls left.
$\displaystyle\sum\limits_{j = 0}^6\dbinom{8-j}{2}=84$ is the number of rows in which the first red ball is followed immediately by a second red ball.
SEE HERE