A Hairy, Ballsy Problem

I don't know if it's right but I came up with a probability of this:
#1,2,3,4 means bag#1,2,3,4
#1 - 10/18 = 55.56%
#2 -  5/18  = 27.78%
#3 -  2/18  = 11.11%
#4 -  1/18  =   5.55%

Make up a grid to show what's in each bag. x means a blue ball, y means a non-blue ball. 

#1 - x x x x x x x x x x
#2 - x x x x x y y y y y
#3 - x x y y y y y y y y
#4 - x y y y y y y y y y

You don't know the numbering of the bag you're using or what you'll pull out. We can assign grid locations to each ball. The last (rightmost) ball for row 1 is an x and it's at location 1,10. The first y on row 3 is at 3,3. The first y on row 4 is at 4,2. 

Now, the probability of selecting any particular grid location is the same as any other grid location. You've picked a ball. What are the chances it's blue? Well, there are 40 equal possibilities and I count 18 blue and 22 non-blue so the chance of getting blue is 18/40 = 45%, hence the chance of getting non-blue is 55%. 

We're told that the ball was blue so how may grid selections are there? 18. There are 10 grid selections available for #1, 5 grid selections for #2 and so on for the other two bags. That's my answer.