Probability of all girls being in same group? There are 10 kids, consisting of 4 girls and 6 boys. The 10 kids are assigned to two different groups of equal size. 
What is the probability that all girls are in the same group? 
To try to solve this problem, I used the binomial distribution.
$${{n}\choose{x}} \cdot p^xq^{n-x}$$
I thought $P(x=4)$ for this problem, so I calculated it as
$${{5}\choose{4}} \cdot {\frac{4}{10}}^x{}\frac{6}{10}^{n-x}$$
I proceeded to multiply this by 2 because there are 2 groups, and thus 2 possibilities of this happening, but I do not think this is the correct approach. Could someone please guide in me the right direction?
 A: Just to put you in the right direction, you need to be using the hypergeometric distribution in this case, instead of the binomial, as you're sampling without replacement.
As an analogy to your problem, assume you have a bowl with 4 white balls and 6 black balls. From it you want to take out 5 balls without putting any back (without replacement). The events you're looking for are:


*

*taking out exactly 4 white balls and 1 black ball;

*taking out 5 black balls (which means that 4 white balls and 1 black ball remain in the bowl).

A: You can surely use the hyper geometric distribution here to get the desired result, however, in this specific question you can approach it in a more intuitive way. 
Part 1: regardless of the number of boys and girls in the group, in how many ways can you form a group of 5 kids? Answer: 10c5 = 252
Part 2: in how many ways you can form a group of 4 girls + 1 boy? Answer: 6 (because there are 6 boys and all 4 girls must be on the team) 
Solution:  6/252 = 0.023
As initially mentioned, you can also use the hyper geometric distribution and you'll have the same result: 
$$\frac{{{4}\choose{4}} * {{6}\choose{1}}}{{10}\choose{5}}$$
A: Alternatively, all girls being in the same group implies all boys are in the same group. Hence:
$$P(\text{all girls})=P(\text{all boys})=P(B_1\cap B_2\cap B_3\cap B_4\cap B_5)=\\
P(B_1)\cdot P(B_2|B_1)\cdot P(B_3|B_1\cap B_2)\cdot P(B_4|B_1\cap B_2\cap B_3)\cdot P(B_5|B_1\cap B_2\cap B_3\cap B_4)=\\
\frac{6}{10}\cdot \frac{5}{9}\cdot \frac{4}{8}\cdot\frac{3}{7}\cdot\frac{2}{6}=\frac{1}{42}\approx 0.0238.$$
