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.

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:
$1/9 + 29/900 + 31/495 = 67/330$
