I'll give this a shot
Afer taking 12 red b***s from the first bag and putting them into the second :
The first bag conains 18 b***s [all red ]
And the second bag now contains :
72 b***s..... 60 black, and 12 red = 5/6 black and 1/6 red
Assuming the second bag is well−mixed and taking 12 b***s from it and putting hem back into the first bag.:
We should select 10 black b***s (5/6)*12 and 2 red ones (1/6)*12 to be put back into the first bag
So the second bag [ theoreically] now contains 50 black and 10 red
And the first bag should now contain 10 black and 20 red
So.......the first bag should contain as many black b***s as the second bag contains red ones !!!