If a group consists of five girls and five boys, what is the probability that all girls will end up on the same team? 
$10$ kids are grouped into an A team with $5$ kids and a B team with five kids.
  If the group consists of five girls and five boys, what is the probability that all girls will end up on the same team

Here is my thought process for this problem:
Choosing first person for Team A, there is a $\frac{5}{10}$ chance girl will be chosen, choosing next person for Team B, $\frac49$ chance of being girl, etc. For simplicity I'll call this $R$.
I believe the solution is $2\cdot R$ since there are two possible teams and either could have the all girl composition
is this correct?
 A: Pick any one of the girls.   She will be on one of the two teams but who cares which.   What we are after is the probability that the other four girls, selected from the other nine players, will also be on her team.

 $$\dfrac{1}{\binom{9}{4}}$$

A: If $R=\frac{5}{10} + \frac{4}{9} + \frac{3}{8} + \frac{2}{7} + \frac{1}{6}$ then the answer is $2R$. 
But this problem is set up to use a hypergeometric distribution. Let $Y=$( # of girls on team A ). In general for a hypergeometric distribtion $P[Y=k]=\dfrac{(_KC_k)(_{(N-K)}C_{(n-k)})}{_{N}C_n}$ . In this case $K$ represents the number of girls total (5 of them) and $N$ is the total amount of people (10 of them) and $k$ is the number of girls we want to select (5 of them) and $n$ is the total amount of people we want to select (5 of them). We want to find the probability that the five girls are on the same team. This is $P[Y=5]+P[Y=0]$ since the five girls can be on team A or team B . So $P[Y=5] + P[Y=0] = \frac{(_5C_5)(_5C_0)}{_{10}C_5} + \frac{(_5C_0)(_5C_5)}{_{10}C_5} = 2 \cdot(.00397) $
A: There are $\frac{10\cdot9\cdot8\cdot7\cdot6}{5\cdot4\cdot3\cdot2\cdot1} = 252$
ways to select team A.  Team B is whoever is not on team A.
Team A could be all all girls or team A could be all boys (making team B all girls) 
$\frac{2}{252} = \frac{1}{126}$
A: I take it that if all the girls are  on team $A \;or\;B,$ the stipulation is met.
The first girl can be placed anywhere, the next has $4$ favorable slots in $9$ remaining, and so on, thus
$Pr = \dfrac49\dfrac38\dfrac27\dfrac16$ 
