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: The idea is to consider the drawn balls as separate groups; namely, look at the draw of the first two, and then look at the last four drawn.
For example, if you were to draw only two balls without replacement, the probability that exactly one is blue is $$\frac{\binom{5}{1}\binom{6}{1}}{\binom{11}{2}}.$$
Now, given that you drew two balls and that one of them was blue, there are now $9$ balls remaining, $4$ of which are blue.  If you draw another four balls, the probability that exactly two are blue is $$\frac{\binom{4}{2} \binom{5}{2}}{\binom{9}{4}}.$$  Therefore, the joint probability of these events is simply $$\frac{\binom{5}{1}\binom{6}{1}\binom{4}{2}\binom{5}{2}}{\binom{11}{2}\binom{9}{4}} = \frac{20}{77}.$$
A: Remember that $Pr(A\cap B)=Pr(A)\times Pr(B\mid A)$.
Let $A$ represent the event "Exactly one of the first two balls is blue" and $B$ the event "Exactly two of the last four balls are blue."
Now., look at the two smaller problems of calculating $Pr(A)$ and $Pr(B\mid A)$

The problem of calculating $Pr(A)$ can be phrased as

A box contains $5$ blue balls and $6$ balls which are not blue.  Two balls are drawn from it.  What is the probability that exactly one of these two balls is blue?

The hypergeometric distribution can indeed be used here.

 $Pr(A)=\dfrac{\binom{5}{1}\binom{6}{1}}{\binom{11}{2}}$


The problem of calculating $Pr(B\mid A)$ can be phrased as

A box contains $4$ blue balls and $5$ balls which are not blue.  Four balls are drawn from it.  What is the probability that exactly two of these four balls are blue?

Use similar techniques as before.
Combine the two answers to arrive at a final answer.
A: Exactly one of the first two is blue 
$\frac {5\cdot 6}{{11\choose2}}$
and exactly 2 of the last 4.
$\frac {{4\choose 2}{5\choose2}}{{9\choose4}}$
and multiply them together.
A: 
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.
