Li'l probability...

2 balls are drawn at random from a bag containing
1 green ball, 1 red ball and 2 blue balls.

If a green ball or a red ball is drawn on 1st draw,
then the ball is put back in the bag; but not if blue.

What is the probability of ending up with a pair
containing 1 blue and 1 green ball, or 1 blue and 1 red ball?

To clarify (in case unclear):

"If a green ball or a red ball is drawn on 1st draw,
then the ball is put back in the bag; but not if blue."

The ball that is put back counts as 1st draw;
like, it doesn't mean "start over"; example:
G is drawn; then put back so we have [G R B B] for 2nd draw;
possible to draw G again, so end up with having drawn 2 G's.

And on this:
"What is the probability of ending up with a pair
containing 1 blue and 1 green ball, or 1 blue and 1 red ball?

Really means: what is probability that the 2 balls
drawn are B and G or B and R (order unimportant)

Not sure about this..

A = the event of drawing red or green, replacing, then blue
B = the event of drawing blue, then red or green

P(A) = (2/4)(2/4) = 1/4
P(B) = (2/4)(2/3) = 1/3

Since A and B are mutually exclusive
P(A or B) = P(A)+P(B) = (1/4)+(1/3) = 7/12

Ball 4! Go to 1st base