2026 AMC 10A Problem 16

There are three jobs: find the lowest possible score, count the draws that get it, and divide by the number of all draws.

Part 1: the lowest possible score.

1. What makes a score low. Every ball scores $1$ point unless its color has already reached its cap: red stops scoring after $3$ balls, black after $2$, white after $2$. So the score is low when many balls are "wasted" past a cap.

1. Organize by the number of red balls. The number of reds can be anything from $0$ to $6$. Once it is chosen, the rest of the $7$ balls are black or white, with at most $4$ of each. To keep the black-and-white points low, pile as many of those balls as possible into one color, so the extras are wasted. For example, with $2$ reds there are $5$ black-and-white balls; putting $4$ in one color scores $2$, and the fifth ball must go in the other color and scores $1$.

(With $1$ red there are $6$ black-and-white balls, and since neither color can have more than $4$, each color gets at least $2$ and scores its full $2$ points.)

3. Result. The lowest possible score is $4$. It happens for exactly four color mixes (red, black, white):
$$(0, 4, 3), \quad (0, 3, 4), \quad (6, 1, 0), \quad (6, 0, 1).$$

Part 2: counting.

4. All draws. Think of the $14$ balls as all different (imagine they are numbered). Then every group of $7$ balls is equally likely, and the number of groups is
$$\binom{14}{7} = \frac{14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 13 \cdot 11 \cdot 2 \cdot 2 \cdot 3 \cdot 2 = 3432.$$
(Cancel $14$ with $7 \cdot 2$, $12$ with $6$, $10$ with $5$, $9$ with $3$, and $8$ with $4$.)

5. Draws with score $4$.
- $(0, 4, 3)$: take all $4$ blacks ($1$ way) and $3$ of the $4$ whites ($4$ ways): $4$ draws.
- $(0, 3, 4)$: the same with black and white swapped: $4$ draws.
- $(6, 1, 0)$: take all $6$ reds ($1$ way) and $1$ of the $4$ blacks ($4$ ways): $4$ draws.
- $(6, 0, 1)$: the same with white instead of black: $4$ draws.

In total $4 + 4 + 4 + 4 = 16$ draws.

6. Probability. Since $3432 = 8 \cdot 429$,
$$\frac{16}{3432} = \frac{2}{429}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{2}{429}}$.

For "lowest possible" questions with several pieces, pick the one quantity that controls the most (here, the number of red balls) and make a table with a row for each value. A table makes it impossible to skip a case, and it shows that the lowest score comes from two very different-looking draws. After that it is ordinary counting: numbering the balls makes all draws equally likely, so the probability is favorable draws over $\binom{14}{7}$.

Find the lowest score by counting wasted balls. The score is $7$ minus the number of wasted balls, so we want to waste as many as possible. A color can only waste balls after it has reached its cap.

• If red is the only color wasting balls: $3$ reds score, up to $3$ more are wasted, and the $7$th ball scores. Lowest score $3 + 1 = 4$, with $6$ reds.
• If black and white both waste balls: they already score $2 + 2 = 4$, so the score is at least $4$, and exactly $4$ only with no reds.
• If red wastes together with black or white: the score is at least $3 + 2 = 5$.
• If only black (or only white) wastes: that color scores $2$ and wastes at most $2$, so at least $3$ of the other balls each score a point. The score is at least $5$.
• If nothing is wasted, the score is $7$.

So the lowest score is $4$, from the same four mixes as before, and the counting finishes the same way.