Let's assume there are 100 balls altogether - 15 red and 85 green. We also need to assume that the partially colour-blind people are told that there are 15 red and 85 green. Thus the instruction to put all the red balls in bag A means that they have to put 15 balls in bag A. (If this assumption is not made....we'll come back to this later!)
So if they are 80% reliable then they will put 80% of 15 red balls (ie12 red balls) in bag A and therefore 3 green balls to total 15.
So p(red ball from bag A) = 12/15 = 0.8.
Now if we don't make the assumption that they know there should be 15 balls put in bag A, they would "see" 80% of the 15 of the red balls as red (ie 12) and also they would see 20% of the 85 green balls as red (ie 17) so they would put 29 balls in bag A. Since only 12 of them were really red, p(red ball from bag A) = 12/29.
I think I'm liking the second scenario!! Nice question.