2025 AMC 10B Problem 14

P11-15~4 minPrint

Basic Probability Basic Counting Set up the equation/formula and compute; no special trick needed Exploit symmetry to reduce work or pair up objects

Nine athletes, no two of whom are the same height, try out for the basketball team. One at a time, they draw a wristband at random, without replacement, from a bag containing $3$ blue bands, $3$ red bands, and $3$ green bands. They are divided into a blue group, a red group, and a green group. The tallest member of each group is named the group captain. What is the probability that the group captains are the three tallest athletes?

Show solution without answering

Key insight

The three tallest are the captains exactly when they land in three different groups. Only their three wristbands matter: 6/8 × 3/7.

Solution

Name the three tallest. Call them $A_1$ , $A_2$ , $A_3$ .

2. When are they the captains?
- If they are in three different groups, each is taller than everyone else in their group (everyone else is shorter than all three). So each one is a captain.
- If two of them share a group, only the taller of those two can be captain, so the captains are not the three tallest.

So we need exactly this: $A_1$ , $A_2$ , $A_3$ get three different colors. What the other six athletes draw does not matter.

3. Think of the three tallest drawing first. Every band is equally likely to go to any athlete, so we can imagine $A_1$ , $A_2$ , $A_3$ drawing before the others.
- $A_1$ draws any band. That leaves $8$ bands: $2$ of $A_1$ 's color and $6$ of the other two colors.
- $A_2$ must get a different color: probability $\frac68$ .
- Now $7$ bands are left, and $3$ of them have the color nobody in the top three has yet. $A_3$ needs one of those: probability $\frac37$ .

4. Multiply.
$\frac68 \times \frac37 = \frac{18}{56} = \frac{9}{28}.$

The answer is $\boxed{\textbf{(C)}\ \frac9{28}}$ .

Why this works

The problem looks like it is about nine people, but only three of them decide the outcome. Once you see that, the question shrinks to "three draws without putting bands back". Because every band is equally likely to reach anyone, you can let the people who matter draw first. Watch for "without replacement": each draw changes what is left in the bag.

Alternative approach

Count all splits. The number of ways to split nine different athletes into a blue three, a red three and a green three is $\frac{9!}{3!\,3!\,3!} = 1680$ . For a good split, give the three tallest three different colors ( $3! = 6$ ways), then put two of the other six in each group ( $\frac{6!}{2!\,2!\,2!} = 90$ ways). That is $6 \times 90 = 540$ good splits, and $\frac{540}{1680} = \frac{9}{28}$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Rank the athletes by height and call the three tallest $A_1, A_2, A_3$ .

The captains are exactly $A_1, A_2, A_3$ if and only if these three are in three different groups. Indeed, if they are in different groups then each is taller than everyone else in its group, so each is a captain; and if two of them share a group, the shorter of the two is not a captain, so the set of captains is not $\{A_1, A_2, A_3\}$ .

So only the colours drawn by the top three athletes matter. The nine bands are $3$ of each colour, drawn without replacement.

$A_1$ draws some band; whatever it is, $6$ of the remaining $8$ bands have a different colour.
Given that $A_2$ has a second colour, $7$ bands remain and exactly $3$ carry the third colour, which is what $A_3$ needs.

Hence
$P = \frac68 \cdot \frac37 = \frac{18}{56} = \frac{9}{28}.$

The answer is $\boxed{\textbf{(C)}\ \frac9{28}}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques