AMC 10 · 2012 · #25
Grade 7 probabilityPick an answer.
AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
A probability over continuous random choices is a volume comparison, so Tool #17 (Visualize Spatial Relationships) is the spine: the triple (x,y,z) is a point in the cube [0,n]³, and the probability is (good volume)/n³. Measuring the good region head-on is messy, so Tool #9 (Solve an Easier Related Problem) fixes one ordering x ＜ y ＜ z by symmetry, turning three absolute-value conditions into two simple gaps. Then Tool #4 (Introduce a Variable) does the key move: a shift substitution that squeezes the gaps out and collapses the good region into a smaller cube, giving its volume in one line. Tool #7 (Identify Subproblems) assembles the clean probability formula, and Tool #3 (Eliminate Possibilities) tests the answer choices against the 1/2 threshold to pin down the smallest n.
Turn the probability into a volume
The point (x,y,z) spreads evenly through the cube [0,n]³, so the probability is the good volume divided by n³.
For an evenly scattered point, "how likely" is just "how much room" — probability is a fraction of volume.
For an evenly scattered point, how likely is just how much room, measured as a share of the volume.
▸ Why?
No position is favoured over another, so the chance is measured by how much of the space works.
▸ Why?
That space is a box, and a box's size is its three lengths multiplied together.
Fix one order to simplify the conditions
By symmetry measure one ordering x ＜ y ＜ z and multiply by 6; the conditions become just y-x ≥ 1, z-y ≥ 1.
Once you line the numbers up smallest-to-largest, "at least 1 apart" is just two neighbor gaps, not three tangled distances.
6.NS.C.7Solve An Easier Related ProblemShift the gaps out with a substitution
Set u=x, v=y-1, w=z-2: the gaps vanish into 0 ≤ u ≤ v ≤ w ≤ n-2, so six such pieces fill a cube of volume (n-2)³.
Sliding each number down by its required gap packs the spread-out points into a solid cube of side n-2.
5.MD.C.5Introduce A VariableWrite the probability and the target inequality
Divide the two volumes: the good-pick probability is ((n-2)/n)³, and we want it to beat 1/2.
Each independent "leave a gap" costs the same factor (n-2)/n, so three of them multiply into a cube.
6.EE.A.1Identify SubproblemsTest the choices against one half
n=9 gives 343/729≈ 0.470, still short; n=10 gives 0.512, over half. The smallest value is 10, choice (D).
Since more room only helps, once a value clears the half-mark every larger one does too — so the first winner is the answer.
6.EE.B.5Eliminate PossibilitiesA probability with random real numbers is really a chunk of volume: the point (x,y,z) fills the cube [0,n]³, the "all gaps ≥ 1" region is a shifted cube of side n-2, so the chance is ((n-2)/n)³ — and it first passes one half at n=10.
- Turn the probability into a volume
- Fix one order to simplify the conditions
- Shift the gaps out with a substitution
- Write the probability and the target inequality
- Test the choices against one half