AMC 10 · 2021 · #19
Grade 8 probabilityPick an answer.
The dice are unknown, so name their face counts a and b and turn every sentence of the problem into a statement about counts of ordered pairs. Because all ab outcomes are equally likely, each probability is (number of pairs with that sum) divided by ab, and the shared denominator makes the first condition a pure counting statement. The key structural fact is that the count for a given sum is capped: only finitely many ordered pairs add to 7, 10, or 12, and the dice being at least 6-sided already guarantees most of them exist. Pushing on those caps squeezes a to a single value, after which the sum-12 condition becomes one linear equation in b. Since we want the least total, we take the smallest surviving pair.
Name the sample space
Every probability shares one denominator.
Fair dice make every ordered pair equally likely, so probability questions become pure counting questions over one shared denominator.
Fair dice make every ordered pair equally likely, so probability questions become pure counting.
▸ Why?
When every outcome carries the same weight, a chance is the favourable count over the total count.
▸ Why?
The two dice are rolled without regard to each other, so every pairing of faces occurs exactly once.
Count the ways to roll seven
There are only six ways.
Seven is small enough that all its pairs use labels 1 through 6, which every die in the problem is guaranteed to have.
7.SP.C.8Make A Systematic ListConvert the ratio into a count
The ratio gives the sum-ten count.
Two probabilities over the same number of outcomes stand in the same ratio as their counts, so the unknown a*b disappears immediately.
6.RP.A.3Introduce A VariableSqueeze the face counts
That count fixes one die's faces.
Nine is the ceiling on ways to make 10, so demanding eight ways says exactly one edge pair is blocked, and only the smaller die can do the blocking.
7.SP.C.8Extreme PrincipleCount the ways to roll twelve
The count splits by the other die's size.
Only the big die's ceiling decides how many of the eight possible partners for 12 actually exist.
7.SP.C.8Make A Systematic ListSolve the equation
Two candidates come out.
Each range of b gives its own formula for n₁₂, so the single probability clue becomes one short linear equation per range.
7.EE.B.4Introduce A VariableTake the smaller and check
The smaller total is 17.
Once the candidate list is down to two dice pairs, the smallest total that survives every check is the answer.
8.EE.C.7Eliminate PossibilitiesWhen dice are fair, probability questions are really counting questions: write every clue as a count of ordered pairs over the same total, and the unknown dice sizes get squeezed out one at a time.
- Name the dice and the sample space
- Count the ways to roll 7
- Convert the ratio into a count
- Squeeze a from the sum-10 cap
- Count the ways to roll 12
- Solve the sum-12 equation
- Take the smaller total and check it