AMC 8 · 2009 · #12
Grade 7 probability
Pick an answer.
AMC 8 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
There are only 3 × 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 × 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.
Two independent spins with 3 sectors each give 3 × 3 = 9 equally likely ordered outcome pairs.
A 3 × 3 grid of outcomes is 3 equal groups of 3 — the Grade 3 meaning of multiplication.
3.OA.A.1Draw A DiagramFill a 3 × 3 table with each row + column sum; the nine sums are 3, 5, 7, 5, 7, 9, 7, 9, 11.
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.
3.OA.A.1Make A Systematic ListAmong the sums only 9 = 3 × 3 is composite, and it fills two cells, so 9 - 2 = 7 cells hold prime sums.
Recognizing 9 = 3 × 3 as composite (and 3, 5, 7, 11 as prime) is Grade 4 factor reasoning.
4.OA.B.4Make A Systematic ListEvery cell of the 3 × 3 grid is equally likely, so the probability is prime cells over total cells = .
With equally likely outcomes, probability is just "how many work" divided by "how many total".
7.SP.C.7Make A Systematic ListThis AMC 8 problem only needs Grade 7 probability — list every outcome on a small grid, count the favorable ones, and divide.