AMC 10 · 2014 · #17
Grade 7 probabilityPick an answer.
AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Because every die is fair and independent, all 6³ = 216 ordered outcomes are equally likely, so the probability is just (favorable outcomes)/216. The phrase "two of the dice sum to the third" is slippery because it does not say which two. Tool #16 (Change Focus) fixes that: decide which single die is the total, then the other two must add up to it. That splits the job into 3 matching subproblems (Tool #7). Inside each subproblem the counting is a clean Tool #2 (Make a Systematic List): list the ordered pairs of the other two dice that add to each reachable total, which fall into the tidy staircase 1, 2, 3, 4, 5.
Count all equally likely outcomes
Three independent fair dice give 216 equally likely ordered outcomes, so the probability is just favorable over 216.
When every outcome is equally likely, probability becomes a counting job: favorable over total.
7.SP.C.7Make A Systematic ListPick which die is the total
Turn it around: first pick which of the three dice is the total — 3 ways — then the other two must add up to it.
Naming which die is the total turns a vague 'some two' into a concrete target to hit.
7.SP.C.8Change Focus Count The ComplementList two-dice sums for one total die
Ordered pairs summing to 2, 3, 4, 5, 6 number 1, 2, 3, 4, 5 — a staircase totalling 15 per choice of total die.
Ordered pairs adding to a target line up as 1, 2, 3, 4, 5 — a staircase that is easy to add.
7.SP.C.8Make A Systematic ListCombine the three cases without overlap
Three choices times 15 gives 45, with no double count: two dice cannot both be the total, or some die would show 0.
Two dice cannot each be the total of the rest at once, so the three cases stay separate.
Two dice cannot each be the total of the rest at once, so the three cases stay separate.
▸ Why?
The cases never overlap, so their counts can simply be added.
▸ Why?
Every roll is just as likely, so the chance is a plain count over all the rolls.
Form the probability
Divide favorable by total: , and 9 cancels to give — choice (D).
Divide the favorable count by the total count once, then reduce the fraction.
7.NS.A.3Identify SubproblemsDecide which die is the total, count the other two's ordered pairs (1+2+3+4+5 = 15), times 3 dice is 45, over 216 gives 5/24.
- Count all equally likely outcomes
- Pick which die is the total
- List two-dice sums for one total die
- Combine the three cases without overlap
- Form the probability