AMC 10 · 2022 · #12
Grade 7 probabilitycountingPick an answer.
Two "at least" conditions at once is the signature of tool #16 (Change Focus / Count the Complement): satisfying them splits into many cases, while failing them splits into very few, so counting the failures and subtracting is the short road. But tool #16 has a sharp edge on this problem and it is worth naming before starting. The failure set of condition 1 and the failure set of condition 2 are not disjoint, so the complement cannot be counted by adding those two sizes. The failures have to be re-cut into pieces that cannot both happen, and choosing that cut is the real work. Before any of that, tool #15 (Organize Information in More Ways) does the move that makes the problem small. Neither condition cares what a die actually reads, only how it sits relative to the thresholds 4 and 2, so the six faces collapse into three blocks — and, by luck of the arithmetic, each block holds exactly two faces and therefore has probability 1/3. Tool #9 (Solve an Easier Related Problem) cashes that in: because the three blocks are equally likely, the 1296 dice outcomes may be replaced by 81 equally likely three-letter patterns with no loss of exactness, and the problem becomes a counting problem small enough to check by hand. Tool #2 (Make a Systematic List) counts the first failure group directly. Tool #3 (Eliminate Possibilities) finishes the second failure group, where knowing one die is high rules out every other die being anything but low.
Group each die into a band
Only three bands matter per die.
When a condition only compares a number against thresholds, the exact face value is noise — keep the band and throw the rest away.
7.SP.C.7Organize Information In More WaysCount the patterns
There are eighty-one equally likely patterns.
Collapsing six faces into three equally likely bands keeps every probability exact while cutting the sample space by a factor of 16.
Collapsing six faces into three equally likely bands keeps every probability exact while shrinking the sample space.
▸ Why?
Each band holds the same number of faces, so each band is just as likely as the others.
▸ Why?
The dice are rolled without regard to each other, so every pattern of bands occurs exactly once.
Cut the failures apart
Split the failures so they cannot overlap.
Two overlapping failure sets turn into two disjoint ones the moment you split on whether condition 1 holds.
7.SP.C.7Change Focus Count The ComplementCount the first failure group
Count the rolls with no high die.
Banning one of three letters leaves two choices per die, so the count is a plain power of two.
7.SP.C.8Make A Systematic ListCount the second failure group
Count the rolls where the high die stands alone.
A die greater than 4 is also greater than 2, so it pays into both conditions at once — which is exactly what forces it to be the only die above 2.
7.SP.C.8Eliminate PossibilitiesSubtract and read the probability
Subtracting gives sixty-one eighty-firsts.
Once every pattern is equally likely, the probability is nothing more than a count of survivors over a count of patterns.
7.SP.C.8Change Focus Count The ComplementSort the six faces into three equally likely groups, then count the rolls that fail instead of the ones that work — but cut the failures so that no roll can land in two groups at once.
- Only each die's band matters
- Eighty-one equally likely patterns
- Count failures, cut so they cannot overlap
- Group 1: no high die at all
- Group 2: the high die stands alone
- Subtract and read the probability