Competition · AMC preparation · step 4 of 4
AMC 8 · 2014 · #12
Grade 7 probabilityPick an answer.
AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
There are only a handful of possible matchings, so Tool #2 (Make a Systematic List) lets us write out every possible guess in order and just count. To make sure the counting rule is right, we first try Tool #9 (Solve an Easier Problem) on the 2-celebrity version — small enough to list by hand — and notice that the count is 2 × 1 = 2. The same idea extends to 3 × 2 × 1 = 6 for the real problem. Once we have the 6 orderings, exactly one is the fully-correct match, so the probability is 1/6.
Try the two-celebrity case
Warm up on the easier 2-celebrity case: listing every pairing gives exactly 2 × 1 = 2 ways.
Trying the small version first checks that 'multiply the choices' gives the same answer as actually listing them.
4.OA.A.3Solve An Easier Related ProblemList all six matchings
List every matching for 3 celebrities — ABC, ACB, BAC, BCA, CAB, CBA — exactly 3 × 2 × 1 = 6 pairings.
An organized list with a fixed ordering rule guarantees we hit every case once and only once.
The three celebrities can be matched to the three baby photos, one photo each, in exactly 3 × 2 × 1 = 6 different ways.
▸ Why?
Whichever photo the first celebrity is given, exactly 2 photos are still free for the second celebrity and then exactly 1 for the third, so the number of open choices at each stage — 3, then 2, then 1 — never depends on the earlier picks.
▸ Why?
Every baby photo belongs to exactly one celebrity and no photo is reused, so the celebrities and the still-unused photos always stay matched one for one; taking one photo out of the pool leaves a pool of exactly the size the count needs.
▸ Why?
Picking a photo for the first celebrity, then the second, then the third are three successive choices, and each celebrity always faces the same number of free photos regardless of what the earlier ones took, so the choices are independent; independent successive choices multiply, so the total is 3 × 2 × 1 = 6.
Count the fully correct matching
Of the 6 orderings, only ABC pairs everyone with their own photo, so there is only 1 fully-correct pairing.
There is only one way to be 'all right,' but many ways to be partly wrong.
7.SP.C.8Make A Systematic ListForm the probability
All 6 pairings are equally likely, so probability = favorable/total = .
When every outcome is equally likely, probability is just 'good cases over all cases.'
7.SP.C.7Make A Systematic ListList all 6 ways to match 3 celebrities to 3 baby photos — only 1 is fully right, so the probability is .
- Try the two-celebrity case
- List all six matchings
- Count the fully correct matching
- Form the probability
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