A pouch holds 4 white and 2 black counters. One counter is drawn at random and is NOT put back; instead, 3 counters of the OTHER colour are dropped into the pouch. A second counter is then drawn at random. What is the probability that this second counter is white?

Split on what the first draw was. If it was white (probability 4/6), the pouch loses a white and gains 3 black, leaving 3 white and 5 black — 8 counters — so the next draw is white with probability 3/8; this branch contributes (4/6) × (3/8) = 12/48. If the first was black (probability 2/6), the pouch loses a black and gains 3 white, leaving 7 white and 1 black, so the next draw is white with probability 7/8; this branch contributes (2/6) × (7/8) = 14/48. Adding the two branches: 12/48 + 14/48 = 26/48 = 13/24. Trap 2/3 reads the pouch as it stood at the start and ignores the top-up altogether. Traps 3/8 and 7/8 each quote a single branch as though the other could not have happened.