Thread: Probability and very bulky solution

1. Probability and very bulky solution

I have a description that there are 9 new balls in the box. I take 3 balls a game and after the game I put these 3 balls back into the box. What's the probability that after the third game there is only one new ball left in the box (there is one ball that has not been used)?
I was trying to use Bayes formula.
FOR THE FIRST GAME:
H1="we take 3 new balls for the first game"
P(H1)=1
FOR THE SECOND GAME:
H2="we take 3 new balls for the second game"
P(H2)=20/84
H3="we take 2 new balls for the second game and 1 old ball"
P(H3)=45/84
H4="we take 1 new ball for the second game and 2 old balls"
P(H4)=18/84
H5="we take 0 new ball for the second game and 3 old balls"
P(H5)=1/84
FOR THE THIRD GAME:
We continue the previous example. The complexity increases for every level.
For example:
We take 3 new balls provided that for the second game we took 3 new balls.
We take 3 new balls provided that for the second game we took 2 new balls.
We take 3 new balls provided that for the second game we took 1 new ball.
We take 3 new balls provided that for the second game we took 0 new ball.
We take 2 new balls provided that for the second game we took 3 new balls.
We take 2 new balls provided that for the second game we took 2 new balls.
and so on .....
It appears to be too bulky solution and gets very messy.
Is my solution towards the solution anything right?
Are there any shorter ways? I wouldn't be able to solve this exercise at class within adequate time.

2. Note after the first game we have three used and six unused balls.
There are just two cases to have exactly one unused ball left.
3N; 3N; 2N & 1U Three new, three new, then one used with 2 new.
3N; 2N & 1U; 3N Three new, one used with 2 new then three new.

It is easy to find the probability of each of the two.