AMC 8 · 2017 · #20
Grade 7 probabilitycountingProblem
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?
Pick an answer.
AMC 8 2017 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.
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?
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) $\tfrac{9}{25}$
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
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 $9999 - 1000 + 1 = 9000$.
💡 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.
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.
💡 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.
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 $10 - 3 = 7$.
💡 Counting how many digits remain after each restriction is repeated subtraction $10 - k$ — well within Grade 4 multi-step arithmetic.
5.NBT.B.5 Step 4 Multiply the choices to get the favorable count.
💡 Multi-digit multiplication like $40 \times 56 = 2240$ is the Grade 5 fluency standard for multiplying multi-digit whole numbers.
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 $40$.
💡 Using prime factorization to find the GCF and reduce a fraction is the Grade 6 GCF/LCM standard.
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 $\tfrac{56}{225}$.
💡 Interpreting a count ratio as a probability between $0$ and $1$ is the Grade 7 introduction to chance probability.
4.NBT.A.2 Subproblem A — count the sample space. The integers from $1000$ to $9999$ inclus 4.OA.A.3 Subproblem B — count favorable outcomes. Let the digits be $d_1 d_2 d_3 d_4$ (th 4.OA.A.3 Count choices for each position. Units $d_4$: $5$ odd digits $\{1,3,5,7,9\}$. Th 5.NBT.B.5 Multiply the choices to get the favorable count. 6.NS.B.4 Subproblem C — form the probability and reduce. Probability $= \dfrac{\text{favo 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.2Read 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.3Solve 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.5Fluently 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.4Find 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.5Understand 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!
⭐ This AMC 8 problem only needs Grade 7 probability — favorable cases divided by total cases — that you already know!
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