$96$ balls in $4$ boxes If there are $72$ white and $24$ red balls to be arranged in $4$ boxes equally (i.e. $24$ balls in each box) what is the probability that all $24$ red balls end up in the same box. 
(If this is a simple question please forgive me, I've lost touch with these kind of problems)
 A: An important point to remember is that arrangements using the stars and bars formula are not equiprobable. A recent answer explaining this can be seen here
Favorable ways are obviously 4,  and total ways = $\binom{96}{24}\;$,  thus $Pr = \dfrac{4}{\binom{96}{24}}$
Further explanation
A lot of comments are being received, so following the dictum "There's more than one way to skin a cat", here's a  colorful explanation that may appeal.
Imagine a line of the 96 balls randomly lined up, and divided by | into 4 boxes
$\color{red}\bullet\circ\circ\circ\color{red}\bullet\circ\circ\circ\circ\circ\circ\circ\color{red}\bullet\circ\circ\circ\circ\circ\circ\circ\color{red}\bullet\circ\circ\circ | \circ ......$
The red balls could be placed in the line in $\binom{96}{24}$ equiprobable ways of which only 4 would have all the red balls in the same box.
A: You can choose $24$ balls for the $1$st box in $\binom{96}{24}$, for the $2$nd box in $\binom{72}{24}$, for the $3$rd box in $\binom{48}{24}$, for the $4$-th box in $\binom{24}{24}=1$ way. If we have to put $24$ red balls in one box we can choose the box to contain them in $4$ ways and fill up the remaining boxes in $\binom{72}{24}\binom{48}{24}\binom{24}{24}$ ways. So the required probability is 
$$
{4\binom{72}{24}\binom{48}{24}\binom{24}{24}\over\binom{96}{24}\binom{72}{24}\binom{48}{24}\binom{24}{24}}={4\over\binom{96}{24}}
$$
like the post of true blue anil says. 
A: The value is the same of the methods by true blue anil and Jack's wasted life but I feel this method easier to understand:
First of let's recognise that placing 24 red balls in 4 buckets is (in this case) the same that placing 24 red balls and 72 white balls in 4 buckets. So I'm evaluating this probability.
We have 4 bucket of capacity 24, so 96 places to place our red balls into.
The first ball can be placed everywhere so its probability is 1 (or 96/96 if you prefer). The second ball, given the position of the first, has only 23 favorable places out of the 95 remaining. The third 22 out of 94 and so on.
So the final probability is: $$
\frac{23}{95}\times\frac{22}{94}\times\frac{21}{93}\times\cdots\frac{1}{73}=\frac{1}{\binom{95}{23}} 
$$
As $96/24=4$: $$
\frac{1}{\binom{95}{23}}=\frac{4}{\binom{95}{23}\frac{96}{24}}=\frac{4}{\binom{96}{24}}
$$
Same value, different ways to reach it.
