Competition · AMC preparation · step 4 of 4
AMC 8 · 2011 · #18
Grade 7 probabilityPick an answer.
AMC 8 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The sample space is the 6 × 6 grid of ordered pairs (a, b), with 36 equally likely outcomes. Tool #2 (Find Symmetry) exploits a clean swap symmetry: the map (a, b) ↦ (b, a) pairs each "first > second" outcome with a unique "first < second" outcome, so those two counts are equal. Tool #7 (Identify Subproblems) splits the event "first ≥ second" into two disjoint cases — strict inequality a > b and equality a = b — that we can count separately and add. Tool #3 (Make a Systematic List) handles the small piece — directly listing the 6 ties (k, k) — and also drives the alternative full-enumeration approach in Review.
Count all outcomes
Each roll is independent and uniform on {1, …, 6}, so the ordered pair (a, b) takes 36 equally likely values.
Listing the sample space of a compound event as ordered pairs is the Grade 7 probability standard for two-stage experiments.
7.SP.C.8Identify SubproblemsSplit the event in two
Split a ≥ b into two disjoint pieces — strict a > b and ties a = b — so P(a ≥ b) = P(a > b) + P(a = b).
Breaking an event into disjoint sub-events is the Tool #7 subproblems move — count each piece, then add.
7.SP.C.8Identify SubproblemsCount the ties
The ordered pairs with a = b are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) — exactly 6 outcomes.
Just listing the diagonal of the 6 × 6 grid uses the Grade 7 uniform-probability counting standard.
7.SP.C.7Eliminate PossibilitiesUse symmetry on the rest
By the swap (a, b) ↦ (b, a), the 30 non-tie pairs split equally between a > b and a < b, giving |{a > b}| = 15.
This is the heart of Tool #2 (Find Symmetry): when a problem treats two roles the same way, swapping them must give equal counts.
Among the 30 rolls where the two numbers differ, exactly 15 have the first roll larger than the second.
▸ Why?
The rolls with the first number larger and the rolls with the first number smaller come in equal counts, so they split the 30 into two matching halves.
▸ Why?
Swapping the two rolls turns each 'first larger' pair into one 'first smaller' pair, and two different pairs never swap to the same result, so the two groups pair off one for one.
▸ Why?
Every differing pair is either 'first larger' or 'first smaller', with none left out and none counted twice, so those two equal groups add back to the full 30.
Add and divide
Add the pieces: 21 favorable outcomes out of 36 give the probability.
Favorable outcomes divided by total outcomes in a uniform sample space is the Grade 7 definition of probability.
7.SP.C.7Identify SubproblemsTwo dice rolls have a built-in symmetry — swapping them shows a > b and a < b happen equally often, so all you have to do is add the ties on top.
- Count all outcomes
- Split the event in two
- Count the ties
- Use symmetry on the rest
- Add and divide
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