AMC 10 · 2023 · #21

Grade 8 probability

probability-basicparitygenerating-functionscombinations-basicpattern-recognition easier-related-problempattern-recognitioncomplementary-counting ↑ Prerequisites: probability-basicparitycombinations-basic

📏 Long solution 💡 4 insights

Problem

Each of $2023$ balls is randomly placed into one of $3$ bins. Which of the following is closest to the probability that each of the bins will contain an odd number of balls?

Pick an answer.

(A)

$frac{2}{3}$

(B)

$frac{3}{10}$

(C)

$frac{1}{2}$

(D)

$frac{1}{3}$

(E)

$frac{1}{4}$

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

Restated: Drop $2023$ distinguishable balls into $3$ bins, each ball uniformly at random and independently. Find the answer choice closest to the probability that all three bins end up with an odd number of balls.

Givens: $2023$ balls, each placed independently into one of $3$ bins with equal probability $\tfrac{1}{3}$; $2023$ is odd; Total bin-counts $X_1 + X_2 + X_3 = 2023$; Answer choices: (A) $\tfrac{2}{3}$, (B) $\tfrac{3}{10}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{1}{3}$, (E) $\tfrac{1}{4}$

Unknowns: The probability $P(\text{all three bin counts are odd})$, then the closest choice

Understand

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #5 Look for a Pattern, #16 Change Focus / Count the Complement, #3 Eliminate Possibilities

We cannot count $3^{2023}$ outcomes directly. Tool #9 (Easier Problem) — replace $2023$ with a small odd number like $n = 1, 3, 5$ — lets us compute the probability by hand. Tool #5 (Pattern) reveals the answer approaches $\tfrac{1}{4}$ very quickly. Tool #16 (Complement) gives the cleanest algebraic confirmation. Finally tool #3 (Eliminate) matches our limit to the closest of the multiple-choice values.

Execute — Answer: E

#9 Solve an Easier Related Problem 2.OA.C.3 Step 1

Replace $2023$ with the smallest odd $n = 1$.
With $1$ ball and $3$ bins, exactly one bin holds $1$ (odd) and the others hold $0$ (even).
All three odd is impossible, so $P_1 = 0$.

$$n = 1: \;P_1 = 0$$

💡 Grade 2 'odd/even' — try the tiniest case to see how parity behaves.

#9 Solve an Easier Related Problem 7.SP.C.8 Step 2

Try $n = 3$.
There are $3^3 = 27$ placements.
The all-odd case forces $(1,1,1)$: pick which ball goes in each bin — that is $3! = 6$ placements.
So $P_3 = \tfrac{6}{27} = \tfrac{2}{9} \approx 0.222$.

$$n=3:\;P_3 = \dfrac{3!}{3^3} = \dfrac{6}{27} = \dfrac{2}{9}$$

💡 Grade 7 'count compound events with an organized list' — small case is fully countable.

#9 Solve an Easier Related Problem 7.SP.C.8 Step 3

Try $n = 5$ to look for the pattern.
The all-odd count partitions of $5$ into three positive odd parts are the permutations of $(1,1,3)$.
Number of ways to assign balls: split $5$ balls into groups of sizes $1, 1, 3$ and assign groups to bins — $\binom{5}{1,1,3}\cdot 3 = \tfrac{5!}{1!\,1!\,3!}\cdot 3 = 20 \cdot 3 = 60$.
So $P_5 = \tfrac{60}{3^5} = \tfrac{60}{243} = \tfrac{20}{81} \approx 0.2469$.

$$n=5:\;P_5 = \dfrac{60}{243} = \dfrac{20}{81} \approx 0.2469$$

💡 Grade 7 — extend the small-case count; watch the probability climb toward a limit.

#5 Look for a Pattern 5.OA.B.3 Step 4

Look at the pattern: $P_1 = 0$, $P_3 \approx 0.222$, $P_5 \approx 0.247$.
The values are rising toward (and falling just short of) $\tfrac{1}{4} = 0.25$.
Conjecture: for large odd $n$, $P_n \to \tfrac{1}{4}$ from below.

$$P_1 = 0,\;P_3 = \tfrac{2}{9} \approx 0.222,\;P_5 = \tfrac{20}{81} \approx 0.2469 \;\nearrow\; \tfrac{1}{4}$$

💡 Grade 5 'analyze patterns and relationships' — the sequence is monotone, heading to a target.

#16 Change Focus / Count the Complement 8.EE.A.1 Step 5

Confirm the $\tfrac{1}{4}$ limit by a symmetry argument (complement-style).
For each ball, mark its bin parity contribution as $+1$ or $-1$ depending on bin choice.
Consider three indicator products.
By the standard generating-trick (or by induction on $n$), the exact probability is $\tfrac{1}{4}\!\left(1 - \tfrac{1}{3^{n-1}}\right)$ for $n$ odd.
With $n = 2023$, the correction term $\tfrac{1}{4\cdot 3^{2022}}$ is essentially zero — astronomically tiny.

$$P_n = \dfrac{1}{4}\!\left(1 - \dfrac{1}{3^{\,n-1}}\right)\;\xrightarrow[n\to\infty]{}\;\dfrac{1}{4}$$

💡 Grade 8 'integer exponents' — the $\tfrac{1}{3^{2022}}$ correction is negligible against $\tfrac{1}{4}$.

#3 Eliminate Possibilities 4.NF.C.7 Step 6

Pick the closest choice.
$P_{2023} \approx \tfrac{1}{4} = 0.25$.
The choices are $\tfrac{2}{3}, \tfrac{3}{10}, \tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{4}$.
The closest is $\tfrac{1}{4}$, choice (E).

$$P_{2023} \approx \tfrac{1}{4} \;\Rightarrow\; \textbf{(E)}$$

💡 Grade 4 'compare decimals' — match $0.25\ldots$ to the answer list.

[1] #9 2.OA.C.3 Replace $2023$ with the smallest odd $n = 1$. With $1$ ball and $3$ bins, exactl

[2] #9 7.SP.C.8 Try $n = 3$. There are $3^3 = 27$ placements. The all-odd case forces $(1,1,1)$:

[3] #9 7.SP.C.8 Try $n = 5$ to look for the pattern. The all-odd count partitions of $5$ into th

[4] #5 5.OA.B.3 Look at the pattern: $P_1 = 0$, $P_3 \approx 0.222$, $P_5 \approx 0.247$. The va

[5] #16 8.EE.A.1 Confirm the $\tfrac{1}{4}$ limit by a symmetry argument (complement-style). For

[6] #3 4.NF.C.7 Pick the closest choice. $P_{2023} \approx \tfrac{1}{4} = 0.25$. The choices are

Review

Reasonableness: Three quick checks pass. (1) The small-case probabilities $0, \tfrac{2}{9}, \tfrac{20}{81}$ all satisfy the formula $\tfrac{1}{4}(1 - 1/3^{n-1})$: for $n=3$, $\tfrac{1}{4}(1 - \tfrac{1}{9}) = \tfrac{1}{4}\cdot\tfrac{8}{9} = \tfrac{2}{9}$ — matches. For $n=5$, $\tfrac{1}{4}(1 - \tfrac{1}{81}) = \tfrac{20}{81}$ — matches. (2) The limit $\tfrac{1}{4}$ also has a heuristic: of the $2^3 = 8$ parity patterns of the three bin counts, only one is (O,O,O), giving a baseline $\tfrac{1}{8}$... but conditioning on the sum being odd halves the space, doubling to $\tfrac{1}{4}$. (3) $0.25$ is closer to all the offered choices than the runners-up $\tfrac{3}{10}=0.3$ (gap $0.05$) and $\tfrac{1}{3}\approx 0.333$ (gap $0.083$).

Alternative: Tool #16 (Change Focus) directly: write each bin count as a sum of $2023$ indicator variables. The probability of all three odd equals the constant term of the expansion of $\big(\tfrac{1}{3}(1 + \omega^a + \omega^{-a})\big)^{2023}$ on parity-roots — the same final formula but reached without the small-case ladder.

CCSS standards used (min grade 8)

2.OA.C.3 Determine whether a group of objects has an odd or even number (Recognizing odd/even bin counts and reasoning that an odd total ($2023$) forbids three-evens patterns.)
4.NF.C.7 Compare two decimals to hundredths by reasoning about their size (Comparing the computed probability ($\approx 0.25$) against each answer choice to pick the closest.)
5.OA.B.3 Generate two numerical patterns using two given rules and identify relationships (Observing the monotone pattern $P_1, P_3, P_5, \ldots$ climbing toward a limit.)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation (Counting favorable placements for the small cases $n=3$ and $n=5$ using organized lists and multinomial counts.)
8.EE.A.1 Know and apply the properties of integer exponents (Recognizing $\tfrac{1}{3^{2022}}$ is negligible and concluding $P_{2023} \approx \tfrac{1}{4}$.)

⭐ This AMC 10 problem only needs Grade 8 exponent reasoning you already know — try $n = 1, 3, 5$ balls instead of $2023$ to see the probability climb $0, \tfrac{2}{9}, \tfrac{20}{81}, \ldots$ toward $\tfrac{1}{4}$. The general formula $\tfrac{1}{4}(1 - 1/3^{n-1})$ has a correction so tiny for $n=2023$ that the closest choice is (E) $\tfrac{1}{4}$.