A tray holds 8 LED bulbs, 3 of which are faulty. A technician tests them one at a time in random order and stops as soon as he has found the second faulty bulb. What is the probability that he stops exactly on the fourth test?

Stopping exactly on the fourth test needs two things at once: exactly ONE faulty bulb among the first three tested, and the fourth bulb faulty. First part: $C(3,1)\times C(5,2)/C(8,3)=(3\times 10)/56=30/56$. Second part: after one faulty and two sound bulbs have gone, 5 bulbs remain of which 2 are faulty, so 2/5. Multiply: $(30/56)\times (2/5)=60/280=3/14$. Trap 15/28 = 30/56 stops at the first part and never asks what the fourth bulb is. Trap 3/8 is just 'the fourth bulb tested is faulty', ignoring the first three. Trap $3/7=C(3,2)\times C(5,2)/C(8,4)$ is the chance that exactly two of the first FOUR bulbs are faulty, which also allows the second faulty one to surface on the second or third test — by then the testing would already have stopped.