AMC 8 · 2017 · #20

Grade 7 probabilitycounting

probability-basicpermutations-basicdigit-constraintsparity systematic-enumerationcasework ↑ Prerequisites: permutations-basicfraction-arithmetic

📏 Medium solution 💡 3 insights

Problem

An integer between 1000 and 9999, inclusive, is chosen at random. What is the probability that it
is an odd integer whose digits are all distinct?

$\textbf{(A) }\frac{14}{75} \qquad \textbf{(B) }\frac{56}{225} \qquad \textbf{(C) }\frac{107}{400} \qquad \textbf{(D) }\frac{7}{25} \qquad \textbf{(E) }\frac{9}{25}$

Pick an answer.

(A)

$frac{14}{75}$

(B)

$frac{56}{225}$

(C)

$frac{107}{400}$

(D)

$frac{7}{25}$

(E)

$frac{9}{25}$

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Toolkit + CCSS Solution

Understand

Restated: Pick a $4$-digit integer at random from $1000$ to $9999$ inclusive. What is the probability that it is odd AND its four digits are all different?

Givens: The integer is chosen uniformly at random from $\{1000, 1001, \ldots, 9999\}$; Total count of $4$-digit integers is $9000$; An integer is odd when its units digit is one of $\{1, 3, 5, 7, 9\}$; "Digits are all distinct" means no digit repeats across the four positions; Answer choices: (A) $\tfrac{14}{75}$, (B) $\tfrac{56}{225}$, (C) $\tfrac{107}{400}$, (D) $\tfrac{7}{25}$, (E) $\tfrac{9}{25}$

Unknowns: Probability $= \dfrac{\text{(odd, all-distinct 4-digit ints)}}{\text{(all 4-digit ints)}}$, in lowest terms

Understand

Restated: Pick a $4$-digit integer at random from $1000$ to $9999$ inclusive. What is the probability that it is odd AND its four digits are all different?

Plan

Primary tool: #7 Identify Subproblems

Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities

Probability problems split cleanly into three subproblems (Tool #7): (a) count the sample space, (b) count the favorable outcomes, (c) form the ratio and reduce. For (b) we count digit by digit using the multiplication principle — a structured systematic list (Tool #2) where the ORDER of filling matters: pick the most restricted slot (units, must be odd) first, then the next-most restricted (thousands, cannot be $0$ and cannot repeat units), then the remaining slots. After computing, Tool #3 (Eliminate) confirms the simplified fraction matches exactly one answer choice.

Execute — Answer: B

#7 Identify Subproblems 4.NBT.A.2 Step 1

Subproblem A — count the sample space.
The integers from $1000$ to $9999$ inclusive are exactly the $4$-digit positive integers; their count is $9999 - 1000 + 1 = 9000$.

$$|\Omega| = 9999 - 1000 + 1 = 9000$$

💡 Knowing that $4$-digit whole numbers run from $1000$ to $9999$ and subtracting to count them is a Grade 4 multi-digit place-value skill.

#2 Make a Systematic List 4.OA.A.3 Step 2

Subproblem B — count favorable outcomes.
Let the digits be $d_1 d_2 d_3 d_4$ (thousands, hundreds, tens, units).
Choose in the order most-restricted-first: pick $d_4$ before $d_1$, because $d_4$ must be odd AND that choice removes a NON-zero digit from the pool, which simplifies $d_1$'s count.

Order of choice: $d_4 \to d_1 \to d_2 \to d_3$

💡 Making a systematic list of digit positions, filled in a fixed order, is a Grade 4 multi-step word-problem strategy using the multiplication principle.

#2 Make a Systematic List 4.OA.A.3 Step 3

Count choices for each position.
Units $d_4$: $5$ odd digits $\{1,3,5,7,9\}$.
Thousands $d_1$: cannot be $0$, cannot equal $d_4$; since $d_4 \ne 0$, exclude both $0$ and $d_4$ from $\{0,\ldots,9\}$, giving $10 - 2 = 8$.
Hundreds $d_2$: any digit not equal to $d_1$ or $d_4$, giving $10 - 2 = 8$.
Tens $d_3$: any digit not equal to $d_1$, $d_2$, $d_4$, giving $10 - 3 = 7$.

$$d_4{:}\,5,\ \ d_1{:}\,8,\ \ d_2{:}\,8,\ \ d_3{:}\,7$$

💡 Counting how many digits remain after each restriction is repeated subtraction $10 - k$ — well within Grade 4 multi-step arithmetic.

#2 Make a Systematic List 5.NBT.B.5 Step 4

Multiply the choices to get the favorable count.

$$5 \times 8 \times 8 \times 7 = 40 \times 56 = 2240$$

💡 Multi-digit multiplication like $40 \times 56 = 2240$ is the Grade 5 fluency standard for multiplying multi-digit whole numbers.

#7 Identify Subproblems 6.NS.B.4 Step 5

Subproblem C — form the probability and reduce.
Probability $= \dfrac{\text{favorable}}{\text{total}} = \dfrac{2240}{9000}$.
Find $\gcd(2240, 9000)$: $2240 = 2^6 \cdot 5 \cdot 7$ and $9000 = 2^3 \cdot 3^2 \cdot 5^3$, so $\gcd = 2^3 \cdot 5 = 40$.
Divide top and bottom by $40$.

$$\dfrac{2240}{9000} = \dfrac{2240 \div 40}{9000 \div 40} = \dfrac{56}{225}$$

💡 Using prime factorization to find the GCF and reduce a fraction is the Grade 6 GCF/LCM standard.

#3 Eliminate Possibilities 7.SP.C.5 Step 6

Interpret as a probability and match to an answer choice (Tool #3).
$\tfrac{56}{225} \approx 0.249$, which is a valid probability between $0$ and $1$.
Scanning the choices, only (B) equals $\tfrac{56}{225}$.

$$P = \dfrac{56}{225} \approx 0.249 \;\Rightarrow\; \textbf{(B)}$$

💡 Interpreting a count ratio as a probability between $0$ and $1$ is the Grade 7 introduction to chance probability.

[1] #7 4.NBT.A.2 Subproblem A — count the sample space. The integers from $1000$ to $9999$ inclus

[2] #2 4.OA.A.3 Subproblem B — count favorable outcomes. Let the digits be $d_1 d_2 d_3 d_4$ (th

[3] #2 4.OA.A.3 Count choices for each position. Units $d_4$: $5$ odd digits $\{1,3,5,7,9\}$. Th

[4] #2 5.NBT.B.5 Multiply the choices to get the favorable count.

[5] #7 6.NS.B.4 Subproblem C — form the probability and reduce. Probability $= \dfrac{\text{favo

[6] #3 7.SP.C.5 Interpret as a probability and match to an answer choice (Tool #3). $\tfrac{56}{

Review

Reasonableness: Two sanity checks. (i) About half of all $4$-digit numbers are odd ($P \approx 0.5$); among those, about $\tfrac{9}{10} \cdot \tfrac{8}{10} \cdot \tfrac{7}{10} \approx 0.504$ have the other three digits distinct from each other and from the units digit — so the joint probability should be around $0.5 \cdot 0.5 \approx 0.25$. Our answer $\tfrac{56}{225} \approx 0.249$ matches this estimate. (ii) The fraction is in lowest terms: $56 = 2^3 \cdot 7$ shares no prime with $225 = 3^2 \cdot 5^2$.

Alternative: Tool #16 (Change Focus / Complement) does NOT help here because both "odd" and "distinct" are restrictive — the complement (even OR has-a-repeat) is messier than the direct count. A cleaner alternative is Tool #9 (Easier Related Problem): first solve the easier problem "how many $4$-digit ints have all distinct digits?" The same multiplication-principle logic gives $9 \cdot 9 \cdot 8 \cdot 7 = 4536$. Restricting further to odd ones, exactly $\tfrac{5}{9}$ of the choices for the units digit work and the constraint that $d_1 \ne 0$ is unaffected, leading to $2240$ again — same answer, useful sanity check.

CCSS standards used (min grade 7)

4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols (Recognizing that the integers $1000$ to $9999$ are exactly the $4$-digit whole numbers and computing the sample-space size $9999 - 1000 + 1 = 9000$.)
4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Working out the per-position digit counts ($5, 8, 8, 7$) by repeatedly subtracting already-used digits from the pool of $10$.)
5.NBT.B.5 Fluently multiply multi-digit whole numbers (Multiplying $5 \times 8 \times 8 \times 7 = 2240$ to combine the per-position digit choices into a total favorable count.)
6.NS.B.4 Find greatest common factor and least common multiple of two numbers (Computing $\gcd(2240, 9000) = 40$ via prime factorization to reduce $\tfrac{2240}{9000}$ to $\tfrac{56}{225}$.)
7.SP.C.5 Understand that the probability of a chance event is between 0 and 1 (Interpreting the simplified count ratio $\tfrac{56}{225}$ as the probability of a chance event (a number between $0$ and $1$) and matching it to the answer choice.)

⭐ This AMC 8 problem only needs Grade 7 probability — favorable cases divided by total cases — that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 7)

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