AMC 10 · 2008 · #20
Grade 7 probabilityThe faces of a cubical die are marked with the numbers 1, 2, 2, 3, 3, and 4. The faces of another die are marked with the numbers 1, 3, 4, 5, 6, and 8. What is the probability that the sum of the top two numbers will be 5, 7, or 9?
Pick an answer.
AMC 10 2008 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: One cube has faces $1, 2, 2, 3, 3, 4$; the other has faces $1, 3, 4, 5, 6, 8$. Roll both and add the two top numbers. Find the probability that this sum is $5$, $7$, or $9$.
Givens: Die A faces: $1, 2, 2, 3, 3, 4$ (note $2$ and $3$ each appear twice); Die B faces: $1, 3, 4, 5, 6, 8$ (all different); Each die is fair: every one of its six faces is equally likely; The two rolls are independent, giving $6 \times 6 = 36$ equally likely face pairs; Choices: (A) $5/18$, (B) $7/18$, (C) $11/18$, (D) $3/4$, (E) $8/9$
Unknowns: The probability that the sum of the two top faces equals $5$, $7$, or $9$
Understand
Restated: One cube has faces $1, 2, 2, 3, 3, 4$; the other has faces $1, 3, 4, 5, 6, 8$. Roll both and add the two top numbers. Find the probability that this sum is $5$, $7$, or $9$.
Givens: Die A faces: $1, 2, 2, 3, 3, 4$ (note $2$ and $3$ each appear twice); Die B faces: $1, 3, 4, 5, 6, 8$ (all different); Each die is fair: every one of its six faces is equally likely; The two rolls are independent, giving $6 \times 6 = 36$ equally likely face pairs; Choices: (A) $5/18$, (B) $7/18$, (C) $11/18$, (D) $3/4$, (E) $8/9$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities, #16 Change Focus / Count the Complement
The clean way through a two-dice problem is a full addition table (tool #2): rows are Die A's six faces, columns are Die B's six faces, and each cell holds the sum. That lays out all $36$ equally likely outcomes at once so nothing is missed or double-counted. Tool #7 (Identify Subproblems) then splits the count into three easy tallies — how many cells equal $5$, how many equal $7$, how many equal $9$ — and adds them. Tool #3 (Eliminate Possibilities) matches the reduced fraction to a choice and rules out the decoys. Tool #16 (Count the Complement) gives a fast independent sanity check via the complementary sums.
Execute — Answer: B
7.SP.C.7 Step 1 Set up the equally likely outcomes
- Each die is fair, so each of its six faces is equally likely, and the two rolls are independent.
- Pairing any face of Die A with any face of Die B gives $6 \times 6 = 36$ outcomes, all with the same probability $\tfrac{1}{36}$.
- Because $2$ and $3$ each appear twice on Die A, they are just more likely to come up — treating the repeated faces as separate outcomes keeps all $36$ pairs equally likely, which is what lets us count.
💡 Fair and independent rolls make every face pair equally likely, so probability is just favorable pairs over $36$.
7.SP.C.8 Step 2 Build the addition table
- Write Die A's faces down the side and Die B's faces across the top, and fill each cell with the sum of its row and column labels.
- This displays all $36$ sums in one grid.
💡 A row-by-column table lists every compound outcome exactly once.
7.SP.C.8 Step 3 Count the favorable cells
- Sweep the table for each target sum separately.
- Sum $= 5$: the cells $1{+}4$, $2{+}3$, $2{+}3$, $4{+}1$ give $4$ cells.
- Sum $= 7$: the cells $1{+}6$, $2{+}5$, $2{+}5$, $3{+}4$, $3{+}4$, $4{+}3$ give $6$ cells.
- Sum $= 9$: the cells $1{+}8$, $3{+}6$, $3{+}6$, $4{+}5$ give $4$ cells.
- Adding the three tallies, the favorable count is $4 + 6 + 4 = 14$.
💡 Breaking one big count into three target-sum counts keeps the tally error-proof.
4.NF.A.1 Step 4 Form and reduce the probability
- The probability is favorable outcomes over total outcomes: $\tfrac{14}{36}$.
- Both numbers share a factor of $2$, so divide top and bottom by $2$ to get $\tfrac{7}{18}$.
💡 Favorable over total gives the probability, and dividing by a common factor keeps its value unchanged.
7.SP.C.5 Step 5 Match the answer choice
- The probability $\tfrac{7}{18}$ is choice (B).
- It is a sensible probability — a number between $0$ and $1$, a bit under $\tfrac{1}{2}$, which fits three target sums out of many.
- The decoy $\tfrac{11}{18}$ is the complement (the chance of NOT getting $5, 7, 9$); $\tfrac{5}{18}$ drops the sum-$7$ cells; and $\tfrac{3}{4}$, $\tfrac{8}{9}$ are far too large.
- The answer is (B).
💡 A valid probability sits between $0$ and $1$, and only $\tfrac{7}{18}$ matches the honest count.
7.SP.C.7 Each die is fair, so each of its six faces is equally likely, and the two rolls 7.SP.C.8 Write Die A's faces down the side and Die B's faces across the top, and fill eac 7.SP.C.8 Sweep the table for each target sum separately. Sum $= 5$: the cells $1{+}4$, $2 4.NF.A.1 The probability is favorable outcomes over total outcomes: $\tfrac{14}{36}$. Bot 7.SP.C.5 The probability $\tfrac{7}{18}$ is choice (B). It is a sensible probability — a Review
Reasonableness: The three tallies $4$, $6$, $4$ sum to $14$ favorable out of $36$, giving $\tfrac{14}{36} = \tfrac{7}{18}$. This is between $0$ and $1$ and slightly below $\tfrac{1}{2}$, which is reasonable for three chosen sums. As a check, notice both dice can only produce sums from $2$ to $12$, and the odd targets $5, 7, 9$ sit right in the busy middle of the range, so a count near a third of the outcomes is believable.
Alternative: Count the complement (tool #16). The possible sums run $2$ through $12$; the ones that are NOT $5, 7, 9$ are $2, 3, 4, 6, 8, 10, 11, 12$. Tally those cells in the table: $2{:}\,1$, $3{:}\,2$, $4{:}\,3$, $6{:}\,4$, $8{:}\,4$, $10{:}\,3$, $11{:}\,2$, $12{:}\,1$, totaling $20$. Then favorable $= 36 - 20 = 14$, so the probability is $\tfrac{14}{36} = \tfrac{7}{18}$ — the same answer (B).
CCSS standards used (min grade 7)
7.SP.C.7Develop a probability model and use it to find probabilities of events (Recognizing all $36$ face pairs as equally likely so probability is favorable outcomes over $36$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Building the row-by-column addition table and counting the $14$ cells whose sum is $5$, $7$, or $9$.)4.NF.A.1Explain why a fraction is equivalent to another fraction (Reducing $\tfrac{14}{36}$ to $\tfrac{7}{18}$ by dividing numerator and denominator by $2$.)7.SP.C.5Understand that probability is a number between 0 and 1 expressing likelihood (Checking $\tfrac{7}{18}$ is a valid probability and matching it to choice (B) over the decoys.)
⭐ Make an addition table of all $36$ face pairs, count the $14$ cells that add to $5$, $7$, or $9$, and the probability is $\tfrac{14}{36} = \tfrac{7}{18}$.
⭐ Make an addition table of all $36$ face pairs, count the $14$ cells that add to $5$, $7$, or $9$, and the probability is $\tfrac{14}{36} = \tfrac{7}{18}$.
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