Does hypergeometric distribution apply in this case? A box contains 4 red balls, 5 blue balls and 2 green balls.
Six balls are drawn without replacement, find the chance that exactly one of the first two is blue AND exactly two of the last four are blue.
The probability that I came up with is:
$\frac {{\binom {2}{1}}\binom {4}{2}}{\binom {6}{3}}$
I feel like this is not correct, as the probability came out to be 60%
I'm kind of wondering if hypergeometric distribution applies in this problem if so how do I properly set up the equation?
Thank you!


A box contains 4 red balls, 5 blue balls and 2 green balls.
  Six balls are drawn without replacement, find the chance that exactly one of the first two is blue AND exactly two of the last four are blue.

The total ways to draw groups of $2$ and $4$ balls from a heap of $4+5+2$ (ie $11$) is $\binom {11}2\binom{11-2}{4}$.   This is also written as $\binom{11}{2,4,5}$ (which is a multinomial coefficient). 
The ways to draw the first two as exactly $1$ from $5$ blue and one from $4+2$ (ie $6$) not blues, and the last four as exactly $2$ from $5-1$ remaining blue and $2$ from $6-1$ remaining not blues is $\binom{5}{1}\binom {6}{1}\binom{5-1}{2}\binom{6-1}{2}$ or $\binom{5}{1,2,2}\binom{6}{1,2,3}$.
The probability is therefore $$\dfrac{\dbinom{5}{1,2,2}\dbinom{6}{1,2,3}}{\dbinom{11}{2,4,5}}$$

Alternatively we may seek the probability for placing the five blue balls in: one from the two first positions, two from the last four, and leaving two among the five unselected, when selecting an arrangement from the eleven balls. $$\dfrac{\dbinom 21\dbinom 4 2\dbinom 52}{\dbinom{11}{5}}$$
These are, of course, equal.
