Solution:

We need a third draw exactly when the first two chips still sum to $4$ or less. The only such pairs are $\{1,2\}$ and $\{1,3\}.$ Each shows up as an ordered pair of first draws in $2$ ways, so there are $4$ favorable sequences out of $5 \cdot 4 = 20$ equally likely ones. The probability is $4/20 = \tfrac15.$ Therefore, the answer is D.