$5$ red balls and $4$ blue balls into 3 boxes Here is a question which was posted by someone and then deleted...
I have 5 red balls and 4 blue balls (assume balls have the same size). Divides balls into 3 different boxes, each box has 3 balls. What is the probability so that there is a least a box which 3 balls in it have the same color.

Here is how I solve it: We have a following possible configurations:
$$ 3r0b+0r3b+2r1b$$
$$ 3r0b+1r2b+1r2b$$
$$ 3r0b+2r1b+0r3b$$

$$ 2r1b+0r3b+3r0b$$
$$ 2r1b+1r2b+2r1b$$
$$ 2r1b+2r1b+1r2b$$
$$ 2r1b+3r0b+0r3b$$

$$ 1r2b+1r2b+3r0b$$
$$ 1r2b+2r1b+1r2b$$
$$ 1r2b+3r0b+1r2b$$

$$ 0r3b+2r1b+3r0b$$
$$ 0r3b+3r0b+2r1b$$
So there is 9 good configuration among 12 and thus the probability is $0,75$. 
Have I missed something? Is there quicker way?
 A: Observe that since each box must have $3$ balls, a configuration is entirely determined by where you put the blue balls.
The stars and bars formula tells you that there are ${6\choose 2}=15$ ways to put the $4$ blue balls in three labelled boxes. Out of these $15$ ways, three are forbidden, where all blue balls are in the same box ($4$ balls in one box is too many). Therefore, $12$ configurations.
Out of these $12$ configurations, the only way to not have three balls of the same color in one box is to have at least one blue in every box (and not three in the same box, but this is automatic since there are four blue).
Once you have put a blue ball in each box (no choice to make here), your only choice is where to put the last one, there are $3$ options. Therefore three out of twelve configurations do not have one box with all balls of the same color.
A: I can't easily convince myself that each of your configurations is equally likely, so it seems to be easier to imagine that each box has 3 distinguishable positions for balls.
Then there are $\binom 95=126$ configurations all in all.
Of these $3\cdot \binom 62=45$ have a box with three red balls.
If there's a box with three blue balls, then there must also be one with three red balls (there are not enough blue balls left over to put one in each of the two remaining boxes), so we have already counted all relevant configurations, and the probability we're after is
$$ \frac{45}{126} = \frac{5}{14}$$
A: I get the same answer as Henning Makholm, but by a different approach.  I imagine starting with the nine balls in a bag, reaching in to grab three that go into the first box, again to grab another three to go into the second box, and then dumping the final three in the third box.
To analyze this, it's convenient to number the nine balls, with 1-4 for the blue balls and 5-9 for the red ones.  If you do, there are 
$${9\choose3}{6\choose3}{3\choose3}=84\cdot20\cdot1=1680$$ 
assignments in total (unlike Henning's $126$). 
Next, it's convenient to count the number of assignments that don't have any boxes with a single color -- which is to say, the number of assignments in which one box has two blue balls and the other two have one blue ball each. This is each to do. The answer is
$$3\cdot{4\choose2}\cdot2\cdot5\cdot{4\choose2}=1080$$
That is, pick which of the $3$ boxes gets two blue balls, then choose $2$ of the $4$ blue balls to go in it, then pick which of the remaining $2$ blue balls goes in the first (say leftmost) box that's still empty, putting the other remaining blue ball in the other empty box, then pick one of the $5$ red balls to go in the box with the pair of blue balls, then choose $2$ of the remaining $4$ red balls to go with the first (leftmost) of the remaining boxes, with the other two going into the other.
The desired probability is thus
$$1-{1080\over1680}=1-{9\over14}={5\over14}$$
A: We can treat the balls of the same color as being distinct (distinguishing them by shade or size or by placing labels on each ball).  
Our sample space consist of all selections in which three balls are placed in each box.  There are $\binom{9}{3}$ ways to select which three balls are placed in the first box, $\binom{6}{3}$ ways to select which three of the remaining six balls are placed in the second box, and $\binom{3}{3}$ ways to place the remaining three balls in the third box.  Hence, there are 
$$\binom{9}{3}\binom{6}{3}\binom{3}{3}$$
ways to distribute the nine balls in such a way that three balls are placed in each box.
There are two ways to distribute the balls so that at least one box is filled with balls of the same color:


*

*Three red balls are placed in one box, three blue balls are placed in a different box, and the remaining three balls are placed in the other box.

*Three red balls are placed in one box and one red and two blue balls are placed in each of the other two boxes.


Three red balls are placed in one box, three blue balls are placed in a different box, and the remaining three balls are placed in the other box:  Choose which of the three boxes will receive three red balls, choose three of the five red balls to go in that box, choose which of the two remaining boxes will receive three blue balls, and choose which three of the four blue balls will go in that box.  The remaining balls must be placed in the other box.  Hence, there are 
$$\binom{3}{1}\binom{5}{3}\binom{2}{1}\binom{4}{3}$$
such distributions.
Three red balls are placed in one box and one red and two blue balls are placed in each of the other boxes:  Choose which of the three boxes receives three red balls, choose which three of the five red balls go in that box, choose which one of the two remaining red balls and which two of the four blue balls go in the leftmost empty box.  The remaining balls must go in the remaining box.  Hence, there are 
$$\binom{3}{1}\binom{5}{3}\binom{2}{1}\binom{4}{2}$$
such distributions.
Hence, the number of favorable cases is 
$$\binom{3}{1}\binom{5}{3}\binom{2}{1}\binom{4}{3} + \binom{3}{1}\binom{5}{3}\binom{2}{1}\binom{4}{2}$$
Therefore, the probability that there is a box that contains three balls of the same color when five red and four blue balls are distributed to three distinct boxes in such a way that three balls are placed in each box is 
$$\frac{\dbinom{3}{1}\dbinom{5}{3}\dbinom{2}{1}\dbinom{4}{3} + \dbinom{3}{1}\dbinom{5}{3}\dbinom{2}{1}\dbinom{4}{2}}{\dbinom{9}{3}\dbinom{6}{3}\dbinom{3}{3}} = \frac{600}{1680} = \frac{5}{14}$$
which agrees with the result obtained by both Barry Cipra and Henning Makholm.
