Competition · AMC preparation · step 4 of 4
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 set of 3 cards
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 total
Count 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 ListKeep the sets topped by 4
Cross 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.
Exactly three of the ten equally likely three-card draws have 4 as their largest value.
▸ Why?
A draw's largest card is 4 exactly when card 4 is taken and card 5 is left out, because 5 is the only card bigger than 4.
▸ Why?
Every draw has one clear top card, so the ten draws split with no overlap into the ones topping out at 3, at 4, and at 5; the max-4 group is precisely the draws that hold 4 but not 5.
▸ Why?
With 4 fixed as the largest, the other two cards must come from the smaller cards {1,2,3}, and there are exactly three such pairs, giving three favorable draws.
▸ Why?
Each favorable draw matches exactly one pair of smaller cards and each pair {1,2},{1,3},{2,3} builds exactly one favorable draw, so the favorable draws are as many as those pairs — three.
Form the probability
Probability = 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!
- List every set of 3 cards
- Count the total
- Keep the sets topped by 4
- Form the probability
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