Competition · AMC preparation · step 4 of 4
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.
Count all nine outcomes
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 in the sum table
Fill 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 ListCheck each sum for primes
Among 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 ListForm the probability
Every 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".
The probability that the two spun numbers add to a prime equals the number of pairings whose sum is prime divided by the total number of equally likely pairings.
▸ Why?
Because every one of the pairings carries the very same amount of chance, the probability of landing in the "prime sum" group is simply what fraction of all the pairings belong to it — the count of prime-sum pairings over the total count of pairings.
▸ Why?
The pairings really are all equally likely, and there are exactly nine of them: each spinner's three sectors are equally likely and one spin does not affect the other, so each of the three first-spin sectors opens into the same three second-spin sectors — three equal groups of three equally likely pairings.
This AMC 8 problem only needs Grade 7 probability — list every outcome on a small grid, count the favorable ones, and divide.
- Count all nine outcomes
- Fill in the sum table
- Check each sum for primes
- Form the probability
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