2018 AMC 10B Problem 6

Three draws are needed when the first two chips do not yet add to more than $4$, and the third chip then pushes the total over $4$.

1. Which first two chips add to $4$ or less? Two different chips from $1$ to $5$:
- $1 + 2 = 3$ (at most $4$)
- $1 + 3 = 4$ (at most $4$)
- every other pair, like $1 + 4 = 5$ or $2 + 3 = 5$, is already more than $4$.

So only the pairs $\{1, 2\}$ and $\{1, 3\}$ keep the drawing going.
2. Does the third draw always stop it? After $\{1,2\}$ the total is $3$ and the smallest chip left is $3$, giving at least $6$. After $\{1,3\}$ the total is $4$ and the smallest chip left is $2$, giving at least $6$. So yes, the third draw always finishes.
3. Count. The first two chips form one of $\binom{5}{2} = 10$ pairs, all equally likely. Two of them work, so the probability is
$$\frac{2}{10} = \frac{1}{5}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{1}{5}}$.

"Exactly three draws" means "not finished after two, finished after three." Here the second part always happens, so the whole question becomes counting pairs with a small sum. Just be sure the favorable cases and the total are counted the same way (both as pairs, or both in order).

Count in order. The first two draws can happen in $5 \times 4 = 20$ orders. The good ones are $(1,2)$, $(2,1)$, $(1,3)$, $(3,1)$, which is $4$ orders. The probability is $\frac{4}{20} = \frac{1}{5}$.

Three draws are required exactly when the first two draws do not already exceed $4$, but the third draw does.

Two distinct chips from $\{1, 2, 3, 4, 5\}$ sum to at most $4$ only for the pairs $\{1, 2\}$ (sum $3$) and $\{1, 3\}$ (sum $4$). In either case the smallest remaining chip is at least $2$, so the third draw pushes the total past $4$ automatically. Hence "three draws" is the same event as "the first two chips are $\{1,2\}$ or $\{1,3\}$."

There are $\binom{5}{2} = 10$ equally likely unordered pairs for the first two chips, and $2$ of them work:
$$\frac{2}{10} = \frac{1}{5}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{1}{5}}$.