2023 AMC 10 B Problem 21

AMC 10 Post-Exam Webinar

Plan Your Next Step: AMC 10, AMC 12, or AIME.

Tue. Nov.17

Save My Spot

AMC 10 Post-Exam Webinar

Plan Your Next Step: AMC 10, AMC 12, or AIME.

Tue. Nov.17

Save My Spot

One Page to Learn. One Point to Earn.

In This Page:

To help you master 2023 AMC 10 B Problem 21, this page includes:

The original AMC problem to try, plus the correct answer

Clear, step-by-step solutions using different methods

Key ideas, useful formulas, and takeaways you can apply to similar problems

Related problems from past AMC exams, plus new practice problems that build the same skills

Want more practice? Scroll down to download Think Academy’s AMC 10 practice packs and keep building your AMC 10 B skills.

2023 AMC10 B Problem 21

Solution

Solution 1

State Transition Recurrence Method

By tracking the probability of having 3 odd bins as balls are added, we can build a recurrence relation. Finding the limit of this recurrence gives us the probability for a very large number of balls.

Recurrence Limit: When a probability sequence is defined by $p_{n + 2} = a \cdot p_{n} + b$ , its long-term behavior can be found by setting $p_{n + 2} = p_{n} = L$ and solving for the limit $L$ .

Step 1: Define the valid parity states

Because the total number of balls is odd, the sum of the balls across the 3 bins is odd. An odd sum of three integers must be formed by either three odd numbers or one odd and two even numbers. Therefore, after placing any odd number of balls, we either have exactly 3 odd bins or exactly 1 odd bin.

P (3 odd bins) = p_{n}

P (1 odd bin) = 1 - p_{n}

Step 2: Analyze adding two balls

To maintain an odd total number of balls, we analyze the transitions when adding exactly 2 balls. Each ball has 3 choices, so there are 9 equally likely ways to place the two new balls.

P (same bin) = \frac{3}{9} = \frac{1}{3}

P (different bins) = \frac{6}{9} = \frac{2}{3}

Step 3: Determine transition probabilities

If we currently have 3 odd bins, placing 2 balls in the same bin flips that bin’s parity twice, leaving all three bins odd. If we currently have 1 odd and 2 even bins, we must place exactly one ball into each of the two even bins to make all three bins odd. There are 2 ways to do this (Ball A in Even 1 and Ball B in Even 2, or vice versa).

P (3 odds \to 3 odds) = \frac{1}{3}

P (1 odd \to 3 odds) = \frac{2}{9}

Step 4: Set up the recurrence relation

The probability of having 3 odd bins after adding two balls is the sum of the probabilities from the two cases above. We can simplify this expression to find a direct relationship between the states.

p_{n + 2} = \frac{1}{3} p_{n} + \frac{2}{9} (1 - p_{n})

p_{n + 2} = \frac{1}{9} p_{n} + \frac{2}{9}

Step 5: Solve for the limit

To find the behavior for a very large number of balls, we assume the probability stabilizes at a limit. We substitute this limit into our recurrence relation and solve. Since 2023 is very large, the exact probability will be extremely close to this limit.

L = \frac{1}{9} L + \frac{2}{9}

\frac{8}{9} L = \frac{2}{9} ⟹ L = \frac{1}{4}

Correct answer: E. $\frac{1}{4}$
The limit is 1/4, which corresponds to answer choice E. We can check small values:

p_{1} = 0

,

p_{3} = 2/9 \approx 0.222

,

p_{5} = 20/81 \approx 0.247

. The sequence rapidly approaches 1/4.

Solution 2

Parity Filter Method

We can use an algebraic expansion to count the exact number of ways to distribute the balls. By adding and subtracting the expansion evaluated at 1 and -1, we can filter out the even exponents and keep only the odd ones.

Parity Filter: To isolate terms with odd exponents in an algebraic expansion, we subtract the expression evaluated at -1 from the expression evaluated at 1, then divide by 2.

Step 1: Set up the counting expression

We want to count the ways to distribute $n = 2023$ distinct balls into 3 distinct bins such that each bin has an odd number of balls. The total number of unrestricted ways is found by setting all variables to 1 in the multinomial expansion.

f (a, b, c) = (a + b + c)^{n} = \sum \frac{n !}{x ! y ! z !} a^{x} b^{y} c^{z}

Total ways = f (1, 1, 1) = 3^{n}

Step 2: Apply the parity filter to one variable

To keep only the terms where the first bin receives an odd number of balls, we evaluate the function at 1 and -1 for the first variable, subtract them, and divide by 2.

\frac{f ( 1 , b , c ) - f ( - 1 , b , c )}{2} = x odd \sum \frac{n !}{x ! y ! z !} 1^{x} b^{y} c^{z}

Step 3: Apply the filter to all three variables

To isolate the terms where all three variables have odd exponents, we apply the parity filter to all three variables simultaneously. This results in a combination of eight evaluations of our function, dividing by 2 for each of the three filters.

\frac{1}{8} [f (1, 1, 1) - f (- 1, 1, 1) - f (1, - 1, 1) + f (- 1, - 1, 1) - f (1, 1, - 1) + f (- 1, 1, - 1) + f (1, - 1, - 1) - f (- 1, - 1, - 1)]

Step 4: Evaluate the eight terms

We evaluate the function for each of the eight sign combinations. Since the total number of balls is odd, any negative base raised to this power will remain negative.

f (1, 1, 1) = (1 + 1 + 1)^{n} = 3^{n}

f (- 1, 1, 1) = (- 1 + 1 + 1)^{n} = 1^{n} = 1

f (1, - 1, 1) = (1 - 1 + 1)^{n} = 1^{n} = 1

f (- 1, - 1, 1) = (- 1 - 1 + 1)^{n} = (- 1)^{n} = - 1

f (1, 1, - 1) = (1 + 1 - 1)^{n} = 1^{n} = 1

f (- 1, 1, - 1) = (- 1 + 1 - 1)^{n} = (- 1)^{n} = - 1

f (1, - 1, - 1) = (1 - 1 - 1)^{n} = (- 1)^{n} = - 1

f (- 1, - 1, - 1) = (- 1 - 1 - 1)^{n} = (- 3)^{n} = - 3^{n}

Step 5: Simplify the filtered expression

We substitute these eight values back into our filtered expression. Notice that the final term changes sign because subtracting a negative yields a positive.

\frac{1}{8} [3^{n} - 1 - 1 + (- 1) - 1 + (- 1) + (- 1) - (- 3^{n})]

\frac{1}{8} [3^{n} - 6 + 3^{n}] = \frac{2 \cdot 3 ^{n} - 6}{8} = \frac{3 ^{n} - 3}{4}

Step 6: Calculate the final probability

The probability is the number of valid distributions divided by the total number of unrestricted distributions. For our large number of balls, the fractional term becomes negligible.

P = \frac{3 ^{n} - 3}{4 \cdot 3 ^{n}} = \frac{1}{4} - \frac{3}{4 \cdot 3 ^{n}}

P \approx \frac{1}{4}

Correct answer: E. $\frac{1}{4}$
The exact probability is

\frac{1}{4} - \frac{1}{4 \cdot 3 ^{2022}}

. This is strictly less than 1/4 but extremely close to it, confirming that 1/4 (Choice E) is the closest answer.

Solution 3

Steady-State Limit — Multiple-Choice Shortcut

Instead of calculating exact probabilities, we can look at the long-term behavior of the system. Because the random placements thoroughly mix the parities, all valid parity states will eventually become equally likely.

Symmetry in Limits: If a random process is symmetric and unrestricted except for a global constraint (like an odd sum), the valid states will become equally likely as the number of steps approaches infinity.

Step 1: Identify the valid parity states

Every ball placed changes the parity of exactly one bin. Because the total number of balls is odd, the sum of the balls across all bins is odd, which strictly limits the possible parity combinations to exactly four valid states.

Sum = Bin_{1} + Bin_{2} + Bin_{3} = 2023

Valid States : (O, O, O), (O, E, E), (E, O, E), (E, E, O)

Step 2: Apply symmetry for large numbers

As the number of balls grows very large, the random placements thoroughly mix the system. Because the transition rules are symmetric, the system does not favor any of the valid states in the long run. Thus, the probabilities approach a uniform distribution among the four states.

P (O, O, O) = P (O, E, E) = P (E, O, E) = P (E, E, O)

P (O, O, O) \to \frac{1}{4}

Correct answer: E. $\frac{1}{4}$
Since there are 4 valid states and they become equally likely, the probability of the (Odd, Odd, Odd) state approaches 1/4. This matches the results from the rigorous methods perfectly.

Adapted Problems to Practice

1 / 10

What Formulas Do You Need for This Problem?

Parity Filter (Roots of Unity Filter):

x odd \sum \frac{n !}{x ! y ! z !} a^{x} b^{y} c^{z} = \frac{f ( a , b , c ) - f ( - a , b , c )}{2} where f (a, b, c) = (a + b + c)^{n}

$n$ is the total power, and $x, y, z$ are the exponents. The alternating signs of $(- 1)^{x}$ cancel out the terms with even exponents and double the terms with odd exponents. Dividing by 2 isolates the exact sum of the odd terms.

Markov Chain Steady State:

In a random process with symmetric transitions among a finite set of valid states, the probability of being in any specific state approaches a uniform distribution as the number of steps grows large.

Applies when the number of steps is very large and the transition rules do not trap the system in a subset of states. The process of adding balls symmetrically mixes the 4 valid parity states, ensuring they become equally likely.

Key Takeaway: Remember This Strategy for Next Time

A process repeating many times (like placing 2023 balls) or a question asking for a probability 'closest to' a value.

Try: Look for a steady-state limit or build a recurrence relation for small steps.

Why: Large numbers often mean the exact value is messy, but the limit is a clean fraction that the system naturally approaches.

A problem requires counting arrangements where quantities must be strictly even or strictly odd.

Try: Use a parity filter by adding and subtracting algebraic expansions evaluated at $1$ and $- 1$ .

Why: The alternating signs of $(- 1)^{k}$ perfectly cancel out the unwanted parities and double the wanted ones, bypassing the need for complex casework.

Unlock More Targeted AMC 10 Practice Resource

Build Stronger Skills. Practice with Purpose. Prepare with Confidence.

Get a complete set of module-focused AMC 10 practice workbooks designed to help students strengthen core concepts, improve problem-solving strategies, and feel more confident on test day.

With 250+ practice problems organized across six essential AMC 10 modules, students can focus on one skill area at a time and learn from clear, step-by-step solutions.

Download the Free AMC 10 Practice Resource Pack

Inside, You’ll Find:

6 module-focused practice workbooks: Algebra, Geometry, Number Theory, Combinatorics, and Probability & Statistics.
250+ AMC 10-style practice problems covering essential concepts and common competition question types.
Complete answer keys and step-by-step solutions to help students understand the reasoning behind each answer.
Organized, print-ready materials for independent study, guided practice, or extra preparation at home.

Get Instant Access: Fill out the form below to receive our exclusive resource pack.

*Please enter a valid email address to access the materials and receive future prep updates.

About Think Academy

Think Academy US Online, part of TAL Education Group (NYSE: TAL), provides structured math and ELA programs that help students build strong academic foundations and grow with confidence. With expert teachers, proven methods, and innovative AI tools, we nurture learning ability, curiosity, and healthy study habits that support long-term growth and a lifelong love of learning.

Check our programs:

Think Kids: Ages 3-5
Math
- Lower Elementary Math — Grades 1–2
- Upper Elementary Math — Grades 3–5
- Middle School Math — Grade 6–8
- High School Math — Grade 9-12
ELA
- Elementary ELA — Grades 2–5

Want more insights on Math & ELA learning and parenting? Fill out the form on this page and subscribe to our newsletter for weekly tips and the latest resources. Join 100,000+ families in our Parent Community!

Categories: Math CompetitionsTags: AMC 10, AMC 10 2023, AMC 10 B 2023

Get expert PreK–12 insights in your inbox every week

To support your child’s learning, confidence, and growth:
✓ Competition & Test Prep
✓ Printable Worksheets
✓ Webinar Invites

Join 100,000+ subscribers in our parent community.