Multi-event probability puzzle - is my answer correct?

[musicgold]

Homework Statement

Is there a more elegant / efficient way to solve this puzzle? See the attached puzzle.

Relevant Equations

At the end of all three 3 moves, the bags should contain:
Bag B: B, B, O, + G/V/Y
Bag A: remaining balls

Here is my attempt.
Beginning state:
Bag B : B, B, O
Bag A : R, R, G, V, Y

Final state:
Bag B: B, B, O, + G/V/Y
Bag A: remaining balls

First possible exchange that would have exactly 3 different colors in each bag is:
Move 1: P ( Arjun moves either the green, violet, or yellow ball to bag B) = 3/5
Move 2. P ( Becca moves one black ball to bag A ) = 2/4
Move 3. P ( Arjun moves the black ball to bag B ) = 1/5
P (exchange 1)= 3/5 x 2/4 x 1/5 = 3/50

Second possible move that would have exactly 3 different colors in each bag is:
Move 1: P ( Arjun moves one red ball to bag B) = 2/5
Move 2. P ( Becca moves the red ball to bag A ) = 1/4
Move 3. P ( Arjun moves either the green, violet, or yellow ball to bag B ) = 3/5
P(exchange 2) = 3/5 x 2/4 x 1/5 = 3/50

P ( exchange 1 OR exchange 2) = 3/50+ 3/50 = 3/25

 

[PeroK]

Are you sure those are the only possibilities?

 

[musicgold]

Are you sure those are the only possibilities?

Ah...

Third possible exchange that would have exactly 3 different colors in each bag is:
Move 1: P ( Arjun moves either the green, violet, or yellow ball to bag B) = 3/5
Move 2. P ( Becca moves the orange ball to bag A ) = 1/4
Move 3. P ( Arjun moves the orange ball to bag B ) = 1/5
P (exchange 3)= 3/5 x 1/4 x 1/5 = 3/100

Fourth possible exchange that would have exactly 3 different colors in each bag is:
Move 1: P ( Arjun moves either G/V/Y ball to bag B) = 3/5
Move 2. P ( Becca moves the G/V/Y ball to bag A ) = 1/4
Move 3. P ( Arjun moves either G/V/Y ball to bag B ) = 3/5
P (exchange 4)= 3/5 x 1/4 x 3/5 = 9/100

P ( exchange 1 OR exchange 2 OR exchange 3 OR exchange 4 ) = 6/100+ 6/100+ 3/100 + 9/100 = 6/25 ?

Any better way of solving this without missing any possible exchanges in the first place?

Thanks

 

[PeroK]

The only idea I have is to note that we have six different colours, with two duplicates. If we have three different colours in a bag, then we must have a d pair of duplicates plus two others. The final configuration must have two B's in one bag and two R's in the other.

As only one ball is taken from bag B, then we must leave the two black balls alone.

Likewise, we either leave the two red balls alone, or move one back and forward.

And, if we leave the red balls alone, it doesn't matter which of the others we choose.

 

[scottdave]

A black ball can move out of bag 2, then back in, if a "non red" ball was moved in first.

 

[haruspex]

There is some symmetry here. You can waste a pair of moves at the start, RR, or a pair at the end, BB. There are three options for the third move.
The only other sequences consist of leaving the reds and blacks alone, 3x2x3.

 

[bob012345]

Ah...

Third possible exchange that would have exactly 3 different colors in each bag is:
Move 1: P ( Arjun moves either the green, violet, or yellow ball to bag B) = 3/5
Move 2. P ( Becca moves the orange ball to bag A ) = 1/4
Move 3. P ( Arjun moves the orange ball to bag B ) = 1/5
P (exchange 3)= 3/5 x 1/4 x 1/5 = 3/100

Fourth possible exchange that would have exactly 3 different colors in each bag is:
Move 1: P ( Arjun moves either G/V/Y ball to bag B) = 3/5
Move 2. P ( Becca moves the G/V/Y ball to bag A ) = 1/4
Move 3. P ( Arjun moves either G/V/Y ball to bag B ) = 3/5
P (exchange 4)= 3/5 x 1/4 x 3/5 = 9/100

P ( exchange 1 OR exchange 2 OR exchange 3 OR exchange 4 ) = 6/100+ 6/100+ 3/100 + 9/100 = 6/25 ?

Any better way of solving this without missing any possible exchanges in the first place?

Thanks

If you are still interested, I think you missed some cases in the third possible exchange.

 

[haruspex]

If you are still interested, I think you missed some cases in the third possible exchange.

Yes, I'm not sure whether @musicgold realised that post #6 leads to a higher number than 6/25.