AMC 10 · 2004 · #11
Grade 7 probabilityPick an answer.
AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Counting every roll where the product beats the sum means chasing a lot of cases. It is far easier to count the opposite — the rolls where the product does NOT beat the sum (product ≤ sum) — and subtract from 64. That is the signature move of Tool #16 (Count the Complement), because the 'bad' rolls turn out to be a small, tidy family. Tool #4 (Introduce a Variable) lets us call the two rolls a and b so we can reason about ab versus a+b in general. Tool #2 (Make a Systematic List) then organizes the few bad rolls into two clean groups: rolls that include a 1, and the single roll (2,2).
Set up the sample space
Call the two top numbers a and b, each from 1 to 8, so there are 64 equally likely ordered rolls.
Two independent choices of 8 things give 8 × 8 equally likely pairs to sort through.
7.SP.C.8Introduce A VariableFlip to the opposite question
Flip it: count the rolls that fail, where ab ≤ a+b, then subtract that small pile from 64.
Counting the small leftover pile and subtracting beats counting the big pile directly.
Counting the small leftover pile and subtracting beats counting the big pile directly.
▸ Why?
Every roll either wins or fails, so the two counts add up to the whole.
▸ Why?
Every pair is just as likely as any other, so the chance is a count over the total count.
Rolls that include a 1
If a die shows 1, the sum wins: 1 · b=b, but 1+b is bigger. That is 8+8-1=15 rolls.
Multiplying by 1 keeps a number the same, but adding 1 makes it bigger, so a 1 always lets the sum win.
6.EE.B.5Make A Systematic ListRolls with both numbers at least 2
With both dice at least 2, only the tie (2,2) fails (2·2=4=2+2); every other pair has the product ahead — 1 bad roll.
Once both dice are 2 or more, multiplying pulls ahead of adding, and (2,2) is the lone break-even point.
6.EE.A.3Make A Systematic ListCount the good rolls and divide
The 15 rolls with a 1 plus the tie (2,2) give 16 failures, so 64-16=48 rolls win: probability 3/4.
Subtract the tiny bad pile from the whole to get the winning count, then turn it into a fraction.
7.SP.C.8Change Focus Count The ComplementCount the easy opposite: the only rolls where the product does not beat the sum are the ones with a 1 (fifteen of them) plus the tie (2,2), so 64-16=48 out of 64 win, giving 3/4.
- Set up the sample space
- Flip to the opposite question
- Rolls that include a 1
- Rolls with both numbers at least 2
- Count the good rolls and divide