What is the probability that the first 2 balls are the same color while the last 2 balls are different colors? A box contains 3 Blue balls, 4 Green balls and 5 Red balls. 4 balls were picked at random without replacement. What is the probability that the first 2 balls are the same color while the last 2 balls are different colors?
What I have tried:
P(B,B,G,R) = 1/99
P(R,R,B,G) = 2/99
P(G,G,R,B) = 1/66
Therefore, the probability of (2 same color and 2 different colors) = (1/99) + (2/99) + (1/66) = 1/22 (Wrong, according to the textbook). 
Please help.
 A: I tried $2$ different methods and got $\frac{67}{330}$ both times. The denominators of the probabilities at each draw are always $12*11*10*9=11880$ so we only have to keep track of the numerators. We also only have to draw one order of the last $2$ balls, then multiply by $2$.
$$\begin{array}{c|c}\text{Config}&\text{Ways}\\\hline
\text{BBBG}&24\\
\text{BBBR}&30\\
\text{BBGR}&120\\
\text{GGGB}&72\\
\text{GGGR}&120\\
\text{GGBR}&180\\
\text{RRRB}&180\\
\text{RRRG}&240\\
\text{RRBG}&240\\
\hline\text{Total}&2412
\end{array}$$
So I get a probability of
$$\frac{2\cdot2412}{11880}=\frac{67}{330}$$
Working out every draw also was the same:
program balls2
   implicit none
   integer i1,i2,i3,j1,j2,j3,j4
   integer total, count
   integer draw(12)
   total = 0
   count = 0
   do i1=1,10
      do i2=i1+1,11
         do i3=i2+1,12
            do j1=1,9
               if(any(j1==[i1,i2,i3])) cycle
               do j2=j1+1,10
                  if(any(j2==[i1,i2,i3])) cycle
                  do j3=j2+1,11
                     if(any(j3==[i1,i2,i3])) cycle
                     do j4=j3+1,12
                        if(any(j4==[i1,i2,i3])) cycle
                        total=total+1
                        draw = 0
                        draw([i1,i2,i3]) = 1
                        draw([j1,j2,j3,j4]) = 4
                        if(any(draw(1)+draw(2)==[1,4,5])) cycle
                        if(any(draw(3)+draw(4)==[0,2,8])) cycle
                        count = count+1
                     end do
                  end do
               end do
            end do
         end do
      end do
   end do
   write(*,*) total,count
end program balls2

Output was
   27720        5628

Which is the same ratio.
A: Your probabilities you calculated for singular events are correct.
in general whenever answering a probability question the order needs to be taken into account as it increases the amount of ways to reach each pattern. To consider all possible often you can use:  Combination wikipedia page
For this question which i take to mean the last 2 balls have to be different to both each other not from the color of the starting 2 balls, it may be easier not to do so.
Considering the first 2 balls only we have
$P(B,B) = 1/22$
$P(G,G) = 1/11$
$P(R,R) = 5/33$
There are only 3 blue balls and thereby we would never be able to have only blue balls picked so lets consider the other possibility that wouldn't work:
$P(B,B,G,G) = 1/22 * 2/15 = 1/165$
$P(B,B,R,R) = 1/22 * 2/9 = 1/99$
Any other combination with $B,B$ at the beginning would work as the last 2 would differ from each other Therefore th probabillity of getting $B,B$ and then it following the rule is $1/22 - 1/165 - 1/99 = 29/990$
Repeat the process with $G,G$:
$P(G,G,G,G) = 1/495,  P(G,G,R,R) = 2/99,P(G,G,B,B) = 1/165$
$\Rightarrow P($Starting in $G,G$ and following rules$) = 1/11 - 1/495 - 2/99 - 1/165 = 31/495$
Again repeat with $R,R$:
$P(R,R,R,R) = 1/99, P(R,R,B,B)= 1/99,P(R,R,G,G) = 2/99$
$\Rightarrow P($Starting in $R,R$ and following rules$) = 5/33 - 1/99 - 1/99 - 2/99 = 1/9$
Now just add the probabilities together and you get:
$1/9 + 29/900 + 31/495 = 67/330$
