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Find the probability in the general case that in between any two red balls are at least two blue balls.

So, at first, I tried to approach the problem by thinking about the probability that no two red balls are drawn consecutively and I found that this probability is

$P(\text{no two red balls drawn consecutively})= \dfrac{\binom{b+1}{r}}{\binom{r+b}{r}}$

where $r$ is the number of red balls and $b$ is the number of blue balls.

I'm not sure how to use this information to solve the problem.

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  • $\begingroup$ subtract this from $1$. $\endgroup$ – user249332 Jan 26 '16 at 16:29
  • $\begingroup$ No two red balls are drawn consecutively can happen even if there is at least one blue ball between red balls. $\endgroup$ – true blue anil Jan 26 '16 at 17:41
  • $\begingroup$ Please describe the precise situation you are talking about. All we are told so far is that there might be red balls and blue balls and drawings somehow. $\endgroup$ – Christian Blatter Jan 26 '16 at 19:13
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Obviously, for the condition to be fulfilled, there must be at least $2(r-1)$ blue balls.

I will illustrate the method with a numerical example. You can easily work out the general formula from it. Suppose $r = 3, b = 6$

Form $3$ blocks with a red ball followed by $2$ blue ones except for the last red ball:

$\uparrow\color{red}\bullet\color{blue}{\bullet\bullet}\;\;\uparrow\color{red}\bullet\color{blue}{\bullet\bullet}\;\;\uparrow\color{red}\bullet\uparrow$

There are $4$ places for the $2$ extra blues at the uparrows

By stars and bars, they can be placed in $\binom{2+4-1}{2} = 10$ ways

Put $x = b - 2(r-1),$ for the extra blues to be placed, in $(r+1)$ compartments

Easy now to work out a general formula for the number of valid arrangements, and the Pr.

Number of valid ways $= \binom{r+x}{x}$

$Pr = \dfrac{\binom{r+x}{x}}{\binom{r+b}{b}}$

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Clearly there are $r+b\choose r$ ways to arrange $r$ red and $b$ blue balls without restriction, so that'll be the denominator for any probability.

For the purpose of this problem, it's convenient to imagine always placing $2$ additional blue balls at the far end of any arrangement. Doing so allows us to reformulate the restriction as saying that each red ball is immediately followed by at least $2$ blue balls. This means we are simply arranging $r$ triplets, each consisting of a red followed by two blue balls, and $b+2-2r$ blue-ball singlets. Thus the desired probability is

$${b+2-r\choose r}\over{r+b\choose r}$$

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