AMC 8 · 2017 · #10
Grade 7 probabilitycountingPick an answer.
AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Only C(5, 3) = 10 possible 3-card sets exist, so we can literally write every one of them down (Tool #2). Once the list is in front of us, Tool #3 (Eliminate) makes it easy to mark which sets have 4 as the largest: any set containing card 5 is out, and any set not containing card 4 is out. Counting what survives gives the probability directly — no combinatorial formulas needed.
List every 3-card group from {1,2,3,4,5} in increasing order — a fixed ordering rule guarantees no duplicates and no gaps.
Writing out all the equally likely outcomes is exactly the "organized list" sample-space move.
7.SP.C.8Make A Systematic ListCount the list — it has 10 groups, so the sample space is 10 equally likely outcomes.
When every outcome is equally likely, the total number of outcomes is the denominator of the probability.
7.SP.C.7Make A Systematic ListCross off any group containing 5 or missing 4; only 3 survive: {1,2,4},{1,3,4},{2,3,4}, each with largest 4.
Eliminating sets that break the "max = 4" rule leaves exactly the favorable ones.
7.SP.C.8Eliminate PossibilitiesProbability = favorable ÷ total = 3 out of 10, which is choice (C).
Counting favorable outcomes and dividing by total outcomes is the definition of probability for equally likely events.
7.SP.C.7Make A Systematic ListThis AMC 8 problem only needs Grade 7 probability with organized lists you already know — write out every way, count the ones that fit, divide!