AMC 10 · 2002 · #24
Grade 7 probabilityPick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The two picks are independent, so the honest way to get the probability is to count outcomes: (number of winning (T,S) situations) ÷ (total situations). Because Tina's sum only takes a handful of values (3 through 9), a systematic list organised by her sum (Tool #2) captures every case without missing or double-counting. For each fixed sum we split off a clean subproblem (Tool #7): how many pairs make that sum, and how many of Sergio's numbers exceed it. Grouping by the value T (Tool #16 — pick the most useful thing to organise around) turns a messy 10 × 10 grid into seven tidy rows we can add up.
Count the total equally-likely outcomes
Tina has C(5, 2)=10 pairs, Sergio has 10 numbers, and the picks are independent — so 100 equally likely outcomes in all.
Independent choices multiply, so the whole sample space is just Tina's count times Sergio's count.
The two picks are made without regard to each other, so the whole sample space is one count times the other.
▸ Why?
Independent choices combine freely, so every pairing of outcomes occurs exactly once.
▸ Why?
Every such pairing is just as likely as any other, so the chance is a count over the total count.
List Tina's pairs by their sum
Sort the 10 pairs by their sum: sums 3 through 9 come from 1,1,2,2,2,1,1 pairs, and those add back to 10.
Sorting the outcomes by their sum guarantees you see each pair once and never twice.
7.SP.C.8Make A Systematic ListCount Sergio's winning numbers per sum
With T fixed, Sergio wins with 10 minus T numbers; times each pair count, the rows give 7, 6, 10, 8, 6, 2, 1.
Once the sum is set, the numbers bigger than it are simply everything above it up to 10.
7.SP.C.7Make A Systematic ListAdd the winning outcomes
Add the rows: 7+6+10+8+6+2+1 = 40, so 40 of the 100 outcomes are wins for Sergio.
The favorable count is just the sum of the per-sum winning counts.
4.OA.A.3Identify SubproblemsForm the probability
Divide favorable by total: 40/100 = 2/5, which is choice (A).
Probability of an equally-likely event is favorable outcomes over total outcomes.
7.SP.C.7Make A Systematic ListGroup Tina's pairs by their sum; for each sum, count how many of Sergio's numbers are bigger, add those up, and divide by all 100 outcomes.
- Count the total equally-likely outcomes
- List Tina's pairs by their sum
- Count Sergio's winning numbers per sum
- Add the winning outcomes
- Form the probability