AMC 8 · 2007 · #24
Grade 7 probabilityProblem
A bag contains four pieces of paper, each labeled with one of the digits 1, 2, 3 or 4, with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of 3?
Pick an answer.
AMC 8 2007 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: A bag holds four slips labeled $1$, $2$, $3$, $4$. Draw three slips one at a time without replacement and read them in the order drawn as a three-digit number. What is the probability that this three-digit number is a multiple of $3$?
Givens: The bag contains the four digits $1$, $2$, $3$, $4$, each appearing exactly once; Three slips are drawn one at a time without replacement; The three drawn digits form a three-digit number in the order they came out; Answer choices: (A) $\tfrac{1}{4}$, (B) $\tfrac{1}{3}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{2}{3}$, (E) $\tfrac{3}{4}$
Unknowns: The probability that the three-digit number is divisible by $3$
Understand
Restated: A bag holds four slips labeled $1$, $2$, $3$, $4$. Draw three slips one at a time without replacement and read them in the order drawn as a three-digit number. What is the probability that this three-digit number is a multiple of $3$?
Givens: The bag contains the four digits $1$, $2$, $3$, $4$, each appearing exactly once; Three slips are drawn one at a time without replacement; The three drawn digits form a three-digit number in the order they came out; Answer choices: (A) $\tfrac{1}{4}$, (B) $\tfrac{1}{3}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{2}{3}$, (E) $\tfrac{3}{4}$
Plan
Primary tool: #16 Change Focus / Count the Right Thing
Secondary: #2 Make a Systematic List
Counting all $4 \times 3 \times 2 = 24$ ordered draws is doable but wasteful, because the digit-sum rule says divisibility by $3$ depends only on which digits are used, not on their order. Tool #16 (Change Focus) shifts the count from $24$ ordered numbers to the $4$ three-element subsets of $\{1,2,3,4\}$ — each subset is equally likely because each contributes the same $6$ orderings. Tool #2 (Systematic List) then enumerates those $4$ subsets and checks each digit sum against the divisibility rule. Two subsets work out of four, so the probability is $\tfrac{2}{4} = \tfrac{1}{2}$.
Execute — Answer: C
4.OA.B.4 Step 1 - Notice that the question only depends on the digit sum.
- By the divisibility-by-$3$ rule, a number is a multiple of $3$ exactly when its digits add to a multiple of $3$.
- Reordering the digits does not change the sum, so it does not change whether the number is a multiple of $3$.
💡 The Grade 4 multiples-and-factors rule says: to test divisibility by $3$, add the digits. Since reshuffling the same three digits keeps the sum, every arrangement of a winning subset is also a winner.
7.SP.C.7 Step 2 - Switch from counting $24$ ordered draws to counting $4$ unordered subsets.
- There are exactly $\binom{4}{3} = 4$ ways to choose which three of the four digits get drawn, and by symmetry each of those subsets is equally likely.
- So the probability becomes (good subsets)$/$(all subsets).
💡 Each three-digit subset arises from the same number of ordered draws ($3! = 6$), so the $24$ ordered outcomes split evenly into $4$ groups of $6$ — and within a group, all six orderings agree on divisibility by $3$.
7.SP.C.8 Step 3 - List the four three-element subsets of $\{1,2,3,4\}$ in order and compute each digit sum.
- A subset is "good" when its sum is divisible by $3$.
💡 Going in order (smallest missing digit: $4, 3, 2, 1$) catches every subset exactly once. Two of the four sums — $6$ and $9$ — are multiples of $3$.
7.SP.C.7 Step 4 - Form the probability.
- Two of the four equally likely subsets produce a multiple of $3$.
💡 Equally likely outcomes mean probability equals the favorable fraction. Half the subsets win, so the answer is one-half.
4.OA.B.4 Notice that the question only depends on the digit sum. By the divisibility-by-$ 7.SP.C.7 Switch from counting $24$ ordered draws to counting $4$ unordered subsets. There 7.SP.C.8 List the four three-element subsets of $\{1,2,3,4\}$ in order and compute each d 7.SP.C.7 Form the probability. Two of the four equally likely subsets produce a multiple Review
Reasonableness: Cross-check by counting ordered numbers directly. Total three-digit numbers: $4 \times 3 \times 2 = 24$. Good subsets are $\{1,2,3\}$ and $\{2,3,4\}$; each contributes $3! = 6$ arrangements, for $6 + 6 = 12$ winning numbers. Probability $= \tfrac{12}{24} = \tfrac{1}{2}$, matching (C). The two counts agree, which confirms that switching to subsets did not change the answer. The result $\tfrac{1}{2}$ is also reasonable: among the four three-digit sums $\{6, 7, 8, 9\}$, exactly two are multiples of $3$, in line with the rough $\tfrac{1}{3}$ density of multiples — boosted here because the digits cluster near $3$ and its multiples.
Alternative: Tool #2 (Systematic List) used directly: write out all $24$ ordered three-digit numbers, divide each by $3$, and count. This is slower but a great sanity check for a younger student — it produces $12$ winners out of $24$. The Tool #16 shortcut just collapses each group of $3! = 6$ arrangements into one subset and counts the four subsets instead.
CCSS standards used (min grade 7)
4.OA.B.4Find factor pairs and multiples; use divisibility ideas (Applying the Grade 4 divisibility-by-$3$ rule: a number is a multiple of $3$ iff its digit sum is a multiple of $3$. This turns the question into a question about digit sums only.)7.SP.C.7Develop a probability model and use it to find probabilities of events (Treating the four three-element subsets of $\{1,2,3,4\}$ as equally likely outcomes (each comes from $6$ ordered draws) and reading the probability as favorable subsets over total subsets.)7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Listing the four subsets $\{1,2,3\}, \{1,2,4\}, \{1,3,4\}, \{2,3,4\}$ systematically and checking each digit sum against the divisibility test.)
⭐ Divisibility by $3$ depends on the digit sum, not the order — so count the four three-digit subsets of $\{1,2,3,4\}$ instead of all $24$ arrangements. Two of the four sums ($6$ and $9$) work, giving probability $\tfrac{1}{2}$.
⭐ Divisibility by $3$ depends on the digit sum, not the order — so count the four three-digit subsets of $\{1,2,3,4\}$ instead of all $24$ arrangements. Two of the four sums ($6$ and $9$) work, giving probability $\tfrac{1}{2}$.
More like this
Same archetype — closest grade level first.