Red and Blue Balls 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.
 A: 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 $= \binom{r+x}{x}$ 
$Pr = \dfrac{\binom{r+x}{x}}{\binom{r+b}{b}}$  
A: 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
$${b+2-r\choose r}\over{r+b\choose r}$$
