A fair coin and a fair die are tossed together. What is the probability of getting a tail on the coin and a multiple of 3 on the die?

The coin and the die do not affect each other, so the sample space has $2\times 6=12$ equally likely outcomes. The favourable ones are T3 and T6, giving 2/12 = 1/6; the same follows from multiplying $(1/2)\times (2/6)=1/6$. Answering 1/3 keeps only the die and forgets the coin. The value 1/12 counts just one outcome, T6, and overlooks that 3 is a multiple of 3 as well. Adding the two probabilities instead of multiplying gives 1/2 + 1/6 = 2/3, which is what 'or' would ask for, not 'and'.