Competition · AMC preparation · step 4 of 4
AMC 10 · 2023A · #7
Grade 7 probabilityPick an answer.
AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Because rolls only add (never subtract), the running total hits 3 on at most one specific roll number. That makes the event split into a few clean cases: "first hit on roll 1", "first hit on roll 2", "first hit on roll 3", "first hit on roll 4". Tool #2 (Systematic List) is built for "list all sequences that…" — list every starting prefix that sums to 3 at exactly that step. Tool #7 (Identify Subproblems) gives the case split, and Tool #3 (Eliminate Possibilities) trims roll 4 before any work — since each roll is ≥ 1, the minimum total after 4 rolls is 4 ＞ 3, so the first-hit-on-roll-4 case is impossible. The remaining sub-probabilities add cleanly because the cases are mutually exclusive.
Rule out the fourth roll
Each roll is at least 1, so after four rolls the total is at least 4 — roll 4 can never be the first hit, leaving only rolls 1, 2, 3.
If you only add positive numbers, the smallest total after k rolls is k — Grade 3 reasoning about repeated addition.
3.OA.D.8Eliminate PossibilitiesCase 1: reach 3 on roll 1
Case 1 — the total is 3 on roll 1 only by rolling a 3 first: one face out of six, so .
One out of six equally likely faces — the Grade 7 "equally likely outcomes" probability model.
7.SP.C.7Make A Systematic ListCase 2: reach 3 on roll 2
Case 2 — first reaches 3 on roll 2 via (1, 2) or (2, 1), neither starting with 3: two ordered pairs give .
Two independent dice rolls give 36 equally likely ordered pairs — Grade 7 compound-event counting.
Two independent rolls give a full grid of equally likely ordered pairs.
▸ Why?
Each roll is made without regard to the other, so the counts multiply.
▸ Why?
Every pair is just as likely, so the chance is a plain count over the whole grid.
Case 3: reach 3 on roll 3
Case 3 — first reaches 3 on roll 3 only via (1, 1, 1), one triple out of 216 equally likely ones, so .
Three independent rolls → 6 × 6 × 6 = 216 outcomes; only (1,1,1) matches — Grade 7 fundamental counting.
7.SP.C.8Make A Systematic ListAdd the three cases
The three cases are mutually exclusive, so add over the common denominator 216: = , choice (B).
Add fractions with unlike denominators by rewriting over the common denominator 216 — Grade 5 fraction addition.
5.NF.A.1Identify SubproblemsThis AMC 10 problem only needs Grade 7 probability of compound events — list every roll prefix that lands exactly on 3, multiply for each roll, and add the cases that can't happen at the same time.
- Rule out the fourth roll
- Case 1: reach 3 on roll 1
- Case 2: reach 3 on roll 2
- Case 3: reach 3 on roll 3
- Add the three cases
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