AMC 8 · 2009 · #12
Grade 7 probabilityProblem
The two spinners shown are spun once and each lands on one of the numbered sectors. What is the probability that the sum of the numbers in the two sectors is prime?
Pick an answer.
AMC 8 2009 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: Two spinners are spun once. The first lands on one of $\{1, 3, 5\}$ with equal chance, and the second lands on one of $\{2, 4, 6\}$ with equal chance. What is the probability that the sum of the two numbers spun is a prime number?
Givens: Spinner 1 has three equal sectors labeled $1, 3, 5$; Spinner 2 has three equal sectors labeled $2, 4, 6$; The two spins are independent and every sector on each spinner is equally likely; Answer choices: (A) $\tfrac{1}{2}$, (B) $\tfrac{2}{3}$, (C) $\tfrac{3}{4}$, (D) $\tfrac{7}{9}$, (E) $\tfrac{5}{6}$
Unknowns: The probability that the sum of the two spun numbers is prime
Understand
Restated: Two spinners are spun once. The first lands on one of $\{1, 3, 5\}$ with equal chance, and the second lands on one of $\{2, 4, 6\}$ with equal chance. What is the probability that the sum of the two numbers spun is a prime number?
Givens: Spinner 1 has three equal sectors labeled $1, 3, 5$; Spinner 2 has three equal sectors labeled $2, 4, 6$; The two spins are independent and every sector on each spinner is equally likely; Answer choices: (A) $\tfrac{1}{2}$, (B) $\tfrac{2}{3}$, (C) $\tfrac{3}{4}$, (D) $\tfrac{7}{9}$, (E) $\tfrac{5}{6}$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #1 Draw a Diagram
There are only $3 \times 3 = 9$ outcome pairs, so the cleanest path is Tool #2 (Make a Systematic List): list every possible sum exactly once with a fixed ordering rule, then count how many are prime. Tool #1 (Draw a Diagram) supports the list by laying it out as a $3 \times 3$ grid (Spinner 1 across the top, Spinner 2 down the side), which makes it visually obvious that no outcome is missed or repeated. With every outcome equally likely, the probability is just (prime sums) $/ 9$.
Execute — Answer: D
3.OA.A.1 Step 1 - Count the total number of equally likely outcomes.
- Spinner 1 has $3$ sectors and Spinner 2 has $3$ sectors, and the two spins are independent, so the number of ordered outcome pairs is $3 \times 3 = 9$.
💡 A $3 \times 3$ grid of outcomes is $3$ equal groups of $3$ — the Grade 3 meaning of multiplication.
3.OA.A.1 Step 2 - List every sum systematically by filling a $3 \times 3$ table: rows are Spinner 2 values $2, 4, 6$, columns are Spinner 1 values $1, 3, 5$, and each cell is the row $+$ column sum.
- The nine sums are $3, 5, 7$ (row $2$), $5, 7, 9$ (row $4$), and $7, 9, 11$ (row $6$).
💡 Sorting by row (Spinner 2) then column (Spinner 1) is the ordering rule that guarantees the list of $9$ sums is complete and has no duplicates.
4.OA.B.4 Step 3 - Check each of the nine sums for primality.
- The distinct sums in the table are $3, 5, 7, 9, 11$.
- Of these, $3, 5, 7, 11$ are prime, and $9 = 3 \times 3$ is composite.
- Cross out only the cells containing $9$: there are two of them (at $(4, 5)$ and $(6, 3)$).
- The remaining $9 - 2 = 7$ cells are prime sums.
💡 Recognizing $9 = 3 \times 3$ as composite (and $3, 5, 7, 11$ as prime) is Grade 4 factor reasoning.
7.SP.C.7 Step 4 - Compute the probability as (favorable outcomes) $/$ (total outcomes).
- Since every cell of the $3 \times 3$ grid is equally likely, the probability is just the count of prime cells over $9$.
💡 With equally likely outcomes, probability is just "how many work" divided by "how many total".
3.OA.A.1 Count the total number of equally likely outcomes. Spinner 1 has $3$ sectors and 3.OA.A.1 List every sum systematically by filling a $3 \times 3$ table: rows are Spinner 4.OA.B.4 Check each of the nine sums for primality. The distinct sums in the table are $3 7.SP.C.7 Compute the probability as (favorable outcomes) $/$ (total outcomes). Since ever Review
Reasonableness: Every sum is odd (odd $+$ even), and the only odd values in range are $3, 5, 7, 9, 11$. Among small odd numbers, only $9$ fails to be prime, so most outcomes should be prime — $\tfrac{7}{9}$ matches that expectation. Compare to the choices: $\tfrac{1}{2}, \tfrac{2}{3}, \tfrac{3}{4}$ each have a denominator that does not divide evenly into $9$ outcomes, so they cannot come from this $3 \times 3$ grid without simplification; $\tfrac{5}{6}$ likewise has the wrong denominator. Only $\tfrac{7}{9}$ fits.
Alternative: Tool #11 (Eliminate Possibilities) on the choices: every outcome is equally likely on a $3 \times 3$ grid, so the probability must be a fraction with denominator dividing $9$ — i.e., $\tfrac{n}{9}$ or $\tfrac{n}{3}$. Of the five choices, only (D) $\tfrac{7}{9}$ has that form (none of $\tfrac{1}{2}, \tfrac{2}{3}, \tfrac{3}{4}, \tfrac{5}{6}$ can be written with denominator $9$ in lowest terms). That alone pins the answer down to (D) before counting any primes.
CCSS standards used (min grade 7)
3.OA.A.1Interpret products of whole numbers (Counting the total $3 \times 3 = 9$ outcome pairs as a rectangular array of (Spinner 1, Spinner 2) cells.)4.OA.B.4Find factor pairs and recognize prime vs. composite numbers in the range 1-100 (Classifying each of the sums $3, 5, 7, 9, 11$ as prime or composite — identifying $9 = 3 \times 3$ as the only composite among them.)7.SP.C.7Develop a probability model and use it to find probabilities of events (Treating the nine $(\text{Spinner 1}, \text{Spinner 2})$ pairs as equally likely outcomes and computing $P(\text{prime sum}) = \tfrac{7}{9}$.)
⭐ This AMC 8 problem only needs Grade 7 probability — list every outcome on a small grid, count the favorable ones, and divide.
⭐ This AMC 8 problem only needs Grade 7 probability — list every outcome on a small grid, count the favorable ones, and divide.
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