AMC 10 · 2005 · #11
Grade 7 probabilityPick an answer.
Every outcome (a pair of bills) is equally likely, so the probability is just (number of good pairs) / (total pairs). Tool #7 (Identify Subproblems) splits the good pairs into two clean cases: pairs that include a 20, and pairs that do not. The first case is easiest counted with Tool #16 (Count the Complement) — count all pairs, then remove the pairs with no20. The second case is a short, finite check handled by Tool #2 (Make a Systematic List): among the remaining bills, which pairs still reach $20?
Count all possible pairs
There are 28 equally likely pairs.
Since every pair of bills is just as likely as any other, counting the pairs is the whole game.
7.SP.C.8Identify SubproblemsAny pair with a $20 already wins
Any pair with the biggest bill already succeeds, giving 13.
It is easier to count the pairs that dodge both twenties and subtract than to list every pair that grabs one.
Counting the pairs that dodge both twenties and subtracting is easier than listing every pair that grabs one.
▸ Why?
Every pair either includes a twenty or it does not, so the two counts add up to the whole.
▸ Why?
Every pair of bills is just as likely as any other, so the chance is a count of pairs over the total count.
Check the pairs with no $20
Among the rest only the two tens reach the target, adding 1.
Without a twenty, the only way to hit $20 is to pair the two biggest remaining bills, the tens.
7.SP.C.8Make A Systematic ListAdd the cases and form the probability
Adding and reducing gives 1/2, choice (D).
Good outcomes over all equally likely outcomes gives the probability, and 14 out of 28 is exactly half.
7.SP.C.7Identify SubproblemsWhen every pick is equally likely, count the winning pairs over all pairs — and split the count into easy cases, like 'has a twenty' versus 'no twenty', so nothing gets missed or double-counted.
- Count all possible pairs
- Any pair with a $20 already wins
- Check the pairs with no $20
- Add the cases and form the probability