chances of a group being all of the same sex I was wondering, if there are 10 girls and 10 boys in a classroom, and they were randomly assigned in groups of four, what are the chances of there being a group with all people inside it the same sex (all boys, for example?)
If possible, give the explanation and the result clearly visible a part from the rest, in percent.
Example, 'there is a 20% chance for 1 group, 5 % for two', etc.
 A: We are dealing with multivariate hypergeometric distribution.
Give the groups numbers $1,2,3,4,5$.
Let $\mathbf{X}=\left(X_{i,j}\right)$ denote a $5\times2$ matrix
of random variables. Here $X_{i,1}$ denotes the number of girls in
group $i$ and $X_{i,2}$ denotes the number of boys in group $i$.
Let $S$ denote the collection of $5\times2$ matrices $\mathbf{x}=\left(x_{i,j}\right)$
such that the $x_{i,j}$ are nonnegative integers that satisfy $x_{i,1}+x_{i,2}=4$
for $i=1,\dots,5$ and $x_{1,j}+\cdots+x_{5,j}=10$ for $j=1,2$.
Then $S$ serves as support of $\mathbf{X}$ with: $$\Pr\left(\mathbf{X}=\mathbf{x}\right)=\frac{\left(10!\right)^{2}\left(4!\right)^{5}}{20!\prod_{i=1}^{5}\prod_{j=1}^{2}x_{i,j}!}$$
for $\mathbf{x}\in S$.
Let $R:=\left\{ \mathbf{x}\in S\mid\exists i,j\; x_{i,j}=0\right\} $.
Then you are asking for $\sum_{\mathbf{x}\in R}\Pr\left(\mathbf{X}=x\right)$.
Quite a job to find this. Uptil now I don't see a shorter route and will be pleasantly surprised if there appears to be one.
A: I take it that at least one same sex group must be there.
It is enough to distribute the girls into $20$ slots, the boys automatically get remaining slots.
Count the unfavorable ways using [ Choose slots for pattern ] $\times$ [ permute ]
$3-3-2-1-1:\; \binom43^2\binom42\binom41^2\frac{5!}{2!2!}$
$3-2-2-2-1:\; \binom43\binom42^3\binom41\frac{5!}{3!} $
$2-2-2-2-2:\; \binom42^5$
Add up to get unfavorable ways $= K,\;\;say$
and $Pr = 1 - \dfrac{K}{\binom{20}{10}}$

ADDED:
If, instead, you want to count exactly one same sex group being there, count the ways for the following patterns. The method has already been explained.
$4-3-1-1-1:$
$4-2-2-1-1:$
$3-3-3-1-0:$
$3-3-2-2-0:$  
