AMC 8 · 2017 · #20

Grade 7 probabilitycounting

Archetype Branching State Enumeration └ Sub-archetype Ordered Placement Count

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

📏 Medium solution 💡 3 insights

Problem

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?

Pick an answer.

(A)

$frac{14}{75}$

(B)

$frac{56}{225}$

(C)

$frac{107}{400}$

(D)

$frac{7}{25}$

(E)

$frac{9}{25}$

AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

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.

1STEP 1

Subproblem A — the sample space. The integers 1000 to 9999 are exactly the 4-digit numbers, so there are 9000 of them.

|Ω| = 9999 - 1000 + 1 = 9000

2STEP 2

Subproblem B — favorable count. Fill the slots most-restricted-first: choose the odd units digit before the thousands digit.

Order of choice: d₄ → d₁ → d₂ → d₃

3STEP 3

Per slot: units d₄ has 5 odd choices; thousands d₁ excludes 0 and d₄ for 8; hundreds d₂ has 8; tens d₃ has 7.

d₄{:} 5, d₁{:} 8, d₂{:} 8, d₃{:} 7

4STEP 4

Multiply the per-slot choices: 5 × 8 × 8 × 7 = 2240 favorable integers.

5 × 8 × 8 × 7 = 40 × 56 = 2240

5STEP 5

Subproblem C — reduce. gcd(2240, 9000) = 40, so $\frac{2240}{9000}$ = $\frac{56}{225}$ .

\frac{2240}{9000}

= (2240 ÷ 40)/(9000 ÷ 40) =

\frac{56}{225}

6STEP 6

$\frac{56}{225}$ ≈ 0.249 is a valid probability, and it matches exactly one choice — (B).

P =

\frac{56}{225}

≈ 0.249 → (B)

Answer

\frac{56}{225}

Two sanity checks. (i) About half of all 4-digit numbers are odd (P ≈ 0.5); among those, about

\frac{9}{10}

\frac{8}{10}

\frac{7}{10}

≈ 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 · 0.5 ≈ 0.25. Our answer

\frac{56}{225}

≈ 0.249 matches this estimate. (ii) The fraction is in lowest terms: 56 = 2³ · 7 shares no prime with 225 = 3² · 5².

💡Key takeaway

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

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