AMC 10 · 2012 · #25
Grade 7 probabilityReal numbers x, y, and z are chosen independently and at random from the interval [0,n] for some positive integer n. The probability that no two of x, y, and z are within 1 unit of each other is greater than 21. What is the smallest possible value of n?
Pick an answer.
AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Three numbers $x$, $y$, $z$ are each picked at random from the interval $[0,n]$, independently. Call a pick "good" if no two of the three numbers land within $1$ unit of one another. As $n$ grows, more room means a good pick becomes more likely. Find the smallest whole number $n$ for which the probability of a good pick is more than $\frac{1}{2}$.
Givens: $x$, $y$, $z$ are chosen independently and uniformly from $[0,n]$; "No two within $1$ unit" means every pairwise distance is at least $1$: $|x-y|\ge 1$, $|y-z|\ge 1$, $|x-z|\ge 1$; $n$ is a positive integer; We need the probability of the good event to exceed $\frac{1}{2}$; Answer choices: (A) $7$, (B) $8$, (C) $9$, (D) $10$, (E) $11$
Unknowns: The smallest positive integer $n$ making the good-pick probability greater than $\frac{1}{2}$
Understand
Restated: Three numbers $x$, $y$, $z$ are each picked at random from the interval $[0,n]$, independently. Call a pick "good" if no two of the three numbers land within $1$ unit of one another. As $n$ grows, more room means a good pick becomes more likely. Find the smallest whole number $n$ for which the probability of a good pick is more than $\frac{1}{2}$.
Givens: $x$, $y$, $z$ are chosen independently and uniformly from $[0,n]$; "No two within $1$ unit" means every pairwise distance is at least $1$: $|x-y|\ge 1$, $|y-z|\ge 1$, $|x-z|\ge 1$; $n$ is a positive integer; We need the probability of the good event to exceed $\frac{1}{2}$; Answer choices: (A) $7$, (B) $8$, (C) $9$, (D) $10$, (E) $11$
Plan
Primary tool: #17 Visualize Spatial Relationships
Secondary: #9 Solve an Easier Related Problem, #4 Introduce a Variable, #7 Identify Subproblems, #3 Eliminate Possibilities
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)$/n^3$. 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 $\frac{1}{2}$ threshold to pin down the smallest $n$.
Execute — Answer: D
7.SP.C.7 Step 1 Turn the probability into a volume
- Because $x$, $y$, $z$ are independent and uniform on $[0,n]$, the point $(x,y,z)$ is spread evenly through the cube $[0,n]^3$, whose volume is $n^3$.
- For a uniform point, the probability of any event equals the volume of the region where the event holds, divided by the total volume $n^3$.
- So the whole problem becomes: measure the volume of the "good" region where all three pairwise gaps are at least $1$, then divide by $n^3$.
💡 For an evenly scattered point, "how likely" is just "how much room" — probability is a fraction of volume.
6.NS.C.7 Step 2 Fix one order to simplify the conditions
- The three variables play identical roles, so the six possible orderings of $x,y,z$ carve the cube into six congruent good pieces.
- Measure just one and multiply by $6$.
- Take the ordering $x<y<z$.
- Then the messy absolute-value conditions collapse: needing $|x-y|\ge 1$ and $|y-z|\ge 1$ becomes simply $y-x\ge 1$ and $z-y\ge 1$ (and these already force $z-x\ge 2$, so the third gap is automatic).
💡 Once you line the numbers up smallest-to-largest, "at least $1$ apart" is just two neighbor gaps, not three tangled distances.
5.MD.C.5 Step 3 Shift the gaps out with a substitution
- Introduce new variables that absorb the required gaps: let $u=x$, $v=y-1$, $w=z-2$.
- Subtracting $1$ from $y$ and $2$ from $z$ exactly uses up the two mandatory unit gaps, so the strict spacing conditions turn into the plain ordering $0\le u\le v\le w\le n-2$.
- That is the set of ordered triples inside a cube of side $n-2$, and an ordered corner-simplex is $\frac{1}{6}$ of its cube.
- So one ordered piece has volume $\frac{(n-2)^3}{6}$, and all six orderings together give the full good volume.
💡 Sliding each number down by its required gap packs the spread-out points into a solid cube of side $n-2$.
6.EE.A.1 Step 4 Write the probability and the target inequality
- Divide the good volume by the cube volume.
- The good region has volume $(n-2)^3$ and the whole cube has volume $n^3$, so the probability of a good pick is $\frac{(n-2)^3}{n^3}$, which is the single ratio $\left(\frac{n-2}{n}\right)^3$ cubed.
- We want this to beat one half.
💡 Each independent "leave a gap" costs the same factor $\frac{n-2}{n}$, so three of them multiply into a cube.
6.EE.B.5 Step 5 Test the choices against one half
- The probability $\left(\frac{n-2}{n}\right)^3$ grows as $n$ grows, so there is one crossover; check the candidates near it.
- At $n=9$: $\left(\frac{7}{9}\right)^3=\frac{343}{729}\approx 0.470$, which is below $\frac{1}{2}$.
- At $n=10$: $\left(\frac{8}{10}\right)^3=\frac{512}{1000}=0.512$, which is above $\frac{1}{2}$.
- So $n=9$ still fails and $n=10$ is the first to succeed.
- The smallest possible value is $n=10$, 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.
7.SP.C.7 Because $x$, $y$, $z$ are independent and uniform on $[0,n]$, the point $(x,y,z) 6.NS.C.7 The three variables play identical roles, so the six possible orderings of $x,y, 5.MD.C.5 Introduce new variables that absorb the required gaps: let $u=x$, $v=y-1$, $w=z- 6.EE.A.1 Divide the good volume by the cube volume. The good region has volume $(n-2)^3$ 6.EE.B.5 The probability $\left(\frac{n-2}{n}\right)^3$ grows as $n$ grows, so there is o Review
Reasonableness: The formula behaves sensibly at the edges: at $n=2$ it gives $0$ (there is no room to keep three points a unit apart), and as $n\to\infty$ it approaches $1$ (almost any pick is fine). In between, $\left(\frac{n-2}{n}\right)^3$ climbs steadily, so the half-mark is crossed exactly once. Our two checks bracket it cleanly: $0.470<0.5<0.512$, so the crossover sits between $9$ and $10$, and $n=10$ is the first integer above it. That $0.512$ only barely clears one half fits a problem placed at the #25 slot, where the threshold is meant to be tight.
Alternative: Instead of testing choices, solve the inequality directly. $\left(\frac{n-2}{n}\right)^3>\frac12$ means $\frac{n-2}{n}>\frac{1}{\sqrt[3]{2}}\approx 0.7937$, i.e. $1-\frac{2}{n}>0.7937$, so $\frac{2}{n}<0.2063$ and $n>\frac{2}{0.2063}\approx 9.69$. The smallest integer above $9.69$ is $10$, matching (D). This confirms the answer without relying on the multiple-choice list.
CCSS standards used (min grade 7)
7.SP.C.7Develop probability models and use them to find probabilities of events (Modeling the uniform random triple as a point in the cube $[0,n]^3$ so that the probability equals the good region's volume divided by $n^3$.)6.NS.C.7Understand ordering and absolute value of rational numbers (Fixing the order $x<y<z$ and rewriting the absolute-value spacing conditions $|x-y|\ge1,\ |y-z|\ge1$ as the neighbor gaps $y-x\ge1$ and $z-y\ge1$.)5.MD.C.5Relate volume to the operations of multiplication and addition (Computing the good region's volume after the substitution collapses it into a cube of side $n-2$, giving $6\cdot\frac{(n-2)^3}{6}=(n-2)^3$.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Assembling the probability as $\frac{(n-2)^3}{n^3}=\left(\frac{n-2}{n}\right)^3$ and forming the inequality against $\frac{1}{2}$.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Testing $n=9$ and $n=10$ against the threshold to find the smallest $n$ for which $\left(\frac{n-2}{n}\right)^3>\frac12$.)
⭐ A 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 $\ge 1$" region is a shifted cube of side $n-2$, so the chance is $\left(\frac{n-2}{n}\right)^3$ — and it first passes one half at $n=10$.
⭐ A 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 $\ge 1$" region is a shifted cube of side $n-2$, so the chance is $\left(\frac{n-2}{n}\right)^3$ — and it first passes one half at $n=10$.
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