AMC 10 · 2016 · #12
Grade 7 probabilityPick an answer.
AMC 10 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The product is even when at least one factor is even, and "at least one" is the classic signal for Tool #16 (Count the Complement): instead of chasing the many ways to get an even product, flip the question and count the one tidy way it fails. A product is odd only when both factors are odd, which is a single, small case. Tool #2 (Make a Systematic List) lets us count the total pairs and the all-odd pairs without missing any. Tool #3 (Eliminate Possibilities) is the final check: only one choice can match the fraction we compute, so the rest are ruled out.
Count all the equally likely pairs
Order doesn't matter, so the number of equally likely pairs is C(5, 2) = 10.
Counting every pair once gives the fair denominator: the total number of equally likely outcomes.
7.SP.C.8Make A Systematic ListFlip to the easier opposite event
Instead of the even case, count the opposite: a product is odd only when both chosen numbers are odd.
An even factor always drags the product to even, so the only escape from "even" is to dodge evens entirely.
7.SP.C.5Count The ComplementCount the all-odd pairs
Only 1, 3, 5 can make an odd product, giving C(3, 2) = 3 all-odd pairs, so P(odd) = 3/10.
Only the three odd numbers can build an odd product, so just count the pairs you can make from those three.
7.SP.C.8Make A Systematic ListSubtract from one to get the answer
Even and odd exhaust all cases, so P(even) = 1 - 3/10 = 7/10 = 0.7, which is choice (D).
Since even and odd are the only two outcomes, knowing one chance hands you the other for free.
7.SP.C.7Eliminate PossibilitiesWhen a question asks for "at least one," count the single way it fails instead and subtract from one.
- Count all the equally likely pairs
- Flip to the easier opposite event
- Count the all-odd pairs
- Subtract from one to get the answer