Competition · AMC preparation · step 4 of 4
AMC 8 · 2007 · #21
Grade 7 probabilityPick an answer.
AMC 8 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Probability with a small deck is a Grade 7 counting question: (favorable outcomes) ÷ (total outcomes). Tool #13 (Count Systematically) fits because we can anchor on the first card and count how many of the remaining 7 cards win with it. That "anchor and count" works for every starting card because of symmetry. As a check, Tool #7 (Split into Subproblems) breaks the winning event into two disjoint cases — "same letter" and "same color" — and adds their probabilities.
Fix the first card
Anchor on the first card: the 7 cards left are all equally likely, so the probability is (winning second cards) ÷ 7.
Grade 7 probability of a uniform model: each of the 7 remaining cards has the same chance, so we just count the good ones.
7.SP.C.7Convert To AlgebraCount the same-color cards
If the first card is red, the other 3 red cards share its color — that's 3 same-color matches.
Each color has 4 cards, so after removing the first card, 3 same-color cards remain.
7.SP.C.8Convert To AlgebraCount the same-letter cards
The only other card with the same letter is the green one — that's 1 same-letter match, and it's a different card.
Each letter appears on exactly 2 cards (one red, one green), so after removing the first card, 1 same-letter card remains.
7.SP.C.8Convert To AlgebraAdd the counts and divide by 7
The two groups never overlap, so total winners = 3 + 1 = 4, and the probability is 4 out of 7.
Grade 7 "probability of a compound event" by listing favorable outcomes — here, the 4 second cards that produce a win.
The probability of a winning pair equals the first card's count of winning partners divided by the seven cards that remain, and that count is its same-color matches added to its same-letter matches.
▸ Why?
Once the first card is set, the second card is equally likely to be any of the seven that remain, so the chance of a win is the count of winning partners among those seven divided by seven.
▸ Why?
A winning partner either shares the first card's color or shares its letter, and these two kinds form separate groups, so the partner count is the size of the color group plus the size of the letter group.
▸ Why?
A single card cannot share both the color and the letter of the first card, since matching both would make it that very card rather than a different partner, so the two groups have no card in common and adding their sizes counts each winner exactly once.
Anchor on the first card, then count the winners among the 7 left: 3 share the color, 1 shares the letter, total 4 out of 7.
- Fix the first card
- Count the same-color cards
- Count the same-letter cards
- Add the counts and divide by 7
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